Wed. Aug 7th, 2024

Jul 16, 2024

# WBCHSE Class 11 Chemistry: Structure of Atom – Chhaya Publication

### MCQs from Structure of Atom

Q1. The set of quantum numbers for the 23rd electron of chromium (atomic number $$=24$$ ) is-
(A) $$3,2,-2,+\frac{1}{2}$$
(B) $$3,1,+1,+\frac{1}{2}$$
(C) $$3,2,+3,+\frac{1}{2}$$
(D) $$3,3,1,+\frac{1}{2}$$

(A) $$3,2,-2,+\frac{1}{2}$$

The set of quantum numbers for the 23rd electron of chromium (atomic number $$= 24$$):

Chromium has an atomic number of 24, meaning it has 24 electrons. Its electronic configuration is:
$$\text{Cr: } (Ar) 3d^5 4s^1$$

To determine the set of quantum numbers for the 23rd electron:
– The 24th electron goes into the $$4s$$ orbital.
– Therefore, the 23rd electron is in the $$3d$$ subshell.

The quantum numbers for an electron in the $$3d$$ subshell are:
– $$n = 3$$ (principal quantum number)
– $$l = 2$$ (azimuthal quantum number for $$d$$ orbitals)
– $$m_l$$ can be $$-2, -1, 0, +1, +2$$ (magnetic quantum number)
– $$m_s = +\frac{1}{2}$$ or $$-\frac{1}{2}$$ (spin quantum number)

Therefore, the set of quantum numbers can be:
$$(3, 2, -2, +\frac{1}{2})$$

$$\boxed{(A) \ 3, 2, -2, +\frac{1}{2}}$$

Q2. In two different H -atoms ( $$A$$ and $$B$$ ) electrons are revolving in orbits of radius $$r$$ and $$4 r$$ respectively. The ratio of the times required for one complete revolution by the electrons in the respective atoms is-
(A) $$1: 4$$
(B) $$1: 2$$
(C) $$1: 8$$
(D) $$2: 1$$

(C) $$1: 8$$

In two different H-atoms (A and B) electrons are revolving in orbits of radius $$r$$ and $$4r$$ respectively. The ratio of the times required for one complete revolution by the electrons in the respective atoms is:

The time period $$T$$ of an electron in orbit is proportional to $$r^{3/2}$$.

For atom A:
$$r_A = r$$

For atom B:
$$r_B = 4r$$

The ratio of the times required for one complete revolution:
$$\frac{T_A}{T_B} = \frac{r_A^{3/2}}{r_B^{3/2}} = \frac{r^{3/2}}{(4r)^{3/2}} = \frac{r^{3/2}}{4^{3/2} r^{3/2}} = \frac{1}{8}$$

$$\boxed{(C) \ 1:8}$$

Q3. The distance traversed by an electron is numerically equal to its de Broglie wavelength. The velocity of the electron is-
(A) $$\sqrt{\frac{h}{m}}$$
(8) $$\sqrt{\frac{m}{h}}$$
(C) $$\sqrt{\frac{h}{p}}$$
(D) $$\sqrt{\frac{h}{2(K E)}}$$

(D) $$\sqrt{\frac{h}{2(K E)}}$$

The distance traversed by an electron is numerically equal to its de Broglie wavelength. The velocity of the electron is:

According to the de Broglie hypothesis:
$$\lambda = \frac{h}{mv}$$

If the distance $$s$$ traversed by the electron is equal to the de Broglie wavelength $$\lambda$$, then:
$$s = \lambda = \frac{h}{mv}$$

We need to express $$v$$ in terms of known quantities. Let’s consider the energy relation:
$$KE = \frac{1}{2}mv^2$$

Rewriting in terms of velocity:
$$v = \sqrt{\frac{2 \cdot KE}{m}}$$

Given that the distance $$s$$ is equal to the de Broglie wavelength:
$$\lambda = s = \frac{h}{mv}$$

Therefore, the correct option for velocity is:
$$\boxed{(D) \ \sqrt{\frac{h}{2(K E)}}}$$

Q4. The uncertainties in the measurement of position and velocity of a particle are equal. So uncertainty in the measurement of its velocity is-
(A) $$\frac{1}{2} \sqrt{\frac{m h}{\pi}}$$
(B) $$\frac{1}{2 \pi} \sqrt{\frac{h}{m}}$$
(C) $$\frac{1}{2} \sqrt{\frac{h}{\pi m}}$$
(D) none of these

(C) $$\frac{1}{2} \sqrt{\frac{h}{\pi m}}$$

The uncertainties in the measurement of position and velocity of a particle are equal. So uncertainty in the measurement of its velocity is:

Using Heisenberg’s uncertainty principle:
$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

Since the uncertainties in position ($$\Delta x$$) and velocity ($$\Delta v$$) are equal:
$$\Delta x = \Delta v$$

Therefore:
$$\Delta x \cdot m \Delta v \geq \frac{h}{4\pi}$$

Given $$\Delta x = \Delta v$$:
$$(\Delta v)^2 \cdot m \geq \frac{h}{4\pi}$$

$$(\Delta v)^2 \geq \frac{h}{4\pi m}$$

$$\Delta v \geq \sqrt{\frac{h}{4\pi m}}$$

$$\boxed{(C) \ \frac{1}{2} \sqrt{\frac{h}{\pi m}}}$$

Q5. The ratio of the energies of the electron of H -atom in the ground state and that of the $$\mathrm{Be}^{3+}$$ ion in the first excited state is-
(A) $$1: 4$$
(B) $$1: 8$$
(C) $$1: 16$$
(D) $$16: 1$$

(A) $$1: 4$$

The ratio of the energies of the electron of H-atom in the ground state and that of the $$\mathrm{Be}^{3+}$$ ion in the first excited state is:

The energy of an electron in a hydrogen-like atom is given by:
$$E_n = -13.6 \frac{Z^2}{n^2}$$

For the H-atom in the ground state ($$Z = 1, n = 1$$):
$$E_H = -13.6 \text{ eV}$$

For the $$\mathrm{Be}^{3+}$$ ion in the first excited state ($$Z = 4, n = 2$$):
$$E_{Be^{3+}} = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4} = -13.6 \times 4 = -54.4 \text{ eV}$$

The ratio of the energies of the H-atom in the ground state to the $$\mathrm{Be}^{3+}$$ ion in the first excited state is:
$$\frac{E_H}{E_{Be^{3+}}} = \frac{-13.6 \text{ eV}}{-54.4 \text{ eV}} = \frac{13.6}{54.4} = \frac{1}{4}$$

$$\boxed{(A) 1:4}$$

Q6. The energy of the electron in the ground state of H atom is -13.6 eV . So energy of the electron revolving in the 5 th orbit of H -atom will be-
(A) $$-0.54 eV$$
(B) $$-5.4 eV$$
(C) $$-0.85 eV$$
(D) $$-2.72 eV$$

(A) $$-0.54 eV$$

The energy of the electron in the ground state of the hydrogen atom is $$-13.6 \text{ eV}$$. So, the energy of the electron revolving in the 5th orbit of the H-atom will be:

The energy of an electron in a hydrogen-like atom is given by:
$$E_n = -\frac{13.6 \text{ eV}}{n^2}$$

For the 5th orbit ($$n = 5$$):
$$E_5 = -\frac{13.6 \text{ eV}}{5^2} = -\frac{13.6}{25} \text{ eV} = -0.544 \text{ eV}$$

$$\boxed{(A) \ -0.54 \text{ eV}}$$

Q7. The uncertainty in the momentum of an electron is $$1.0 \times 10^{-5} \mathrm{~kg} \mathrm{~ms}^{-1}$$. The uncertainty in its position will be $$\left(h=6.625 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}\right)$$ –
(A) $$1.05 \times 10^{-28} \mathrm{~m}$$
(B) $$1.05 \times 10^{-26} \mathrm{~m}$$
(C) $$5.27 \times 10^{-30} \mathrm{~m}$$
(D) $$5.25 \times 10^{-28} \mathrm{~m}$$

(C) $$5.27 \times 10^{-30} \mathrm{~m}$$

The uncertainty in the momentum of an electron is $$1.0 \times 10^{-5} \text{ kg m/s}$$. The uncertainty in its position is calculated using Heisenberg’s uncertainty principle:

$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$

Given:
$$\Delta p = 1.0 \times 10^{-5} \text{ kg m/s}$$
$$h = 6.625 \times 10^{-34} \text{ kg m}^2/\text{s}$$

$$\Delta x \geq \frac{h}{4\pi \Delta p} = \frac{6.625 \times 10^{-34}}{4\pi \times 1.0 \times 10^{-5}}$$

$$\Delta x \geq \frac{6.625 \times 10^{-34}}{4 \times 3.14159 \times 10^{-5}}$$

$$\Delta x \geq \frac{6.625 \times 10^{-34}}{12.56636 \times 10^{-5}}$$

$$\Delta x \geq \frac{6.625 \times 10^{-34}}{1.256636 \times 10^{-4}}$$

$$\Delta x \geq 5.27 \times 10^{-30} \text{ m}$$

$$\boxed{(C) \ 5.27 \times 10^{-30} \text{ m}}$$

Q8. In $$\mathrm{He}^{+}$$ion the distance of the electron from the nucleus is-
(A) 21.4 pm
(B) 52.9 pm
(C) 26.5 pm
(D) 105.8 pm

(C) 26.5 pm

In $$\mathrm{He}^{+}$$ ion the distance of the electron from the nucleus:

The Bohr radius $$a_0$$ for hydrogen is 52.9 pm. For a hydrogen-like ion with atomic number $$Z$$, the radius of the $$n$$-th orbit is given by:
$$r_n = \frac{n^2 a_0}{Z}$$

For $$\mathrm{He}^{+}$$ ($$Z = 2$$), in the ground state ($$n = 1$$):
$$r_1 = \frac{1^2 \times 52.9 \text{ pm}}{2} = \frac{52.9}{2} \text{ pm} = 26.45 \text{ pm}$$

$$\boxed{(C) \ 26.5 \text{ pm}}$$

Q9. The ratio between kinetic energy and the total energy of the electron of hydrogen atom according to Bohr’s model is-
(A) $$2: 1$$
(B) $$1: 1$$
(C) $$1:-1$$
(D) $$1: 2$$

(A) $$2: 1$$

The ratio between kinetic energy and the total energy of the electron of hydrogen atom according to Bohr’s model:

In Bohr’s model, the total energy $$E$$ is given by:
$$E = -\frac{K}{2}$$

Where $$K$$ is the kinetic energy. Hence,
$$K = -2E$$

Thus, the ratio of kinetic energy to total energy is:
$$\frac{K}{E} = \frac{-2E}{E} = -2$$

$$\boxed{(A) \ 2:1}$$

Q10. The wavelength of the photons of a radiation of wave number $$8 \times 10^{15} \mathrm{~s}^{-1}$$ is $$\left(h=6.625 \times 10^{-34} \mathrm{Is} ; c=3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)-$$
(A) $$3 \times 10^{7} \mathrm{~nm}$$
(B) $$2 \times 10^{25} \mathrm{~nm}$$
(C) $$3 \times 10^{-18} \mathrm{~nm}$$
(D) $$4 \times 10^{1} \mathrm{~nm}$$

(D) $$4 \times 10^{1} \mathrm{~nm}$$

The wavelength of the photons of a radiation of wave number $$8 \times 10^{15} \text{ s}^{-1}$$.

Wave number ($$\nu$$) is typically given in $$\text{m}^{-1}$$ but here, it seems like the given value is in $$\text{s}^{-1}$$, which is the frequency.

The formula to find the wavelength ($$\lambda$$) from frequency ($$\nu$$) is:
$$\lambda = \frac{c}{\nu}$$

Given:
$$\nu = 8 \times 10^{15} \text{ s}^{-1}$$
$$c = 3.0 \times 10^8 \text{ m/s}$$

Calculate $$\lambda$$:
$$\lambda = \frac{3.0 \times 10^8 \text{ m/s}}{8 \times 10^{15} \text{ s}^{-1}}$$
$$\lambda = \frac{3.0}{8} \times 10^{-7} \text{ m}$$
$$\lambda = 0.375 \times 10^{-7} \text{ m}$$
$$\lambda = 3.75 \times 10^{-8} \text{ m}$$
$$\lambda = 37.5 \text{ nm}$$

Given the options, the correct answer is closest to:
$$\boxed{(D) \ 4 \times 10^{1} \text{ nm}}$$

Q11. Which of the following ions has the highest value of magnetic moment-
(A) $$\mathrm{Mn}^{2+}$$
(B) $$\mathrm{Sc}^{3+}$$
(C) $$\mathrm{Ti}^{3+}$$
(D) $$\mathrm{Cr}^{2+}$$

(A) $$\mathrm{Mn}^{2+}$$

The magnetic moment ($$\mu$$) for an ion is given by the formula:
$$\mu = \sqrt{n(n+2)} \text{ BM}$$
where $$n$$ is the number of unpaired electrons.

Let’s determine the number of unpaired electrons for each ion:

$$\mathrm{Mn}^{2+}$$: Mn has an atomic number of 25. The electron configuration is $$(Ar) 3d^5 4s^2$$. For $$\mathrm{Mn}^{2+}$$, we lose the 4s electrons and have 3d^5, which has 5 unpaired electrons.
$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$$

$$\mathrm{Sc}^{3+}$$: Sc has an atomic number of 21. The electron configuration is $$(Ar) 3d^1 4s^2$$. For $$\mathrm{Sc}^{3+}$$, we lose all the 3d and 4s electrons, leaving no unpaired electrons.
$$\mu = \sqrt{0(0+2)} = 0 \text{ BM}$$

$$\mathrm{Ti}^{3+}$$: Ti has an atomic number of 22. The electron configuration is $$(Ar) 3d^2 4s^2$$. For $$\mathrm{Ti}^{3+}$$, we lose the 4s electrons and one 3d electron, leaving one unpaired electron.
$$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$$

$$\mathrm{Cr}^{2+}$$: Cr has an atomic number of 24. The electron configuration is $$(Ar) 3d^5 4s^1$$. For $$\mathrm{Cr}^{2+}$$, we lose the 4s and one 3d electron, leaving 4 unpaired electrons.
$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$$

So, $$\mathrm{Mn}^{2+}$$ has the highest magnetic moment.

$$\boxed{(A) \ \mathrm{Mn}^{2+}}$$

Q12. The angular momentum of the electron in the first excited state in H -atom $$\mathrm{is}-$$
(A) zero
(B) $$\sqrt{2(2+1)} \frac{h}{2 \pi}$$
(C) $$\frac{h}{2 \pi}$$
(D) $$\frac{h}{\pi}$$

(D) $$\frac{h}{\pi}$$

The angular momentum of the electron in the first excited state in H-atom is:

The angular momentum $$L$$ of an electron in a hydrogen atom is given by:
$$L = n \frac{h}{2\pi}$$

For the first excited state ($$n = 2$$):
$$L = 2 \frac{h}{2\pi} = \frac{h}{\pi}$$

$$\boxed{(D) \ \frac{h}{\pi}}$$

Q13. If the wavelength of the line having least value in the Lyman series of hydrogen spectrum be $$x$$, then the wavelength of the line having highest value in the Balmer series of $$\mathrm{Li}^{2+}$$ ion will be-
(A) $$\frac{x}{9}$$
(B) $$\frac{4 x}{5}$$
(C) $$\frac{5 x}{4}$$
(D) $$\frac{4 x}{9}$$

(D) $$\frac{4 x}{9}$$

If the wavelength of the line having least value in the Lyman series of hydrogen spectrum be $$x$$, then the wavelength of the line having highest value in the Balmer series of $$\mathrm{Li}^{2+}$$ ion will be:

For the Lyman series in hydrogen, the least wavelength corresponds to the transition from $$n = \infty$$ to $$n = 1$$. For the Balmer series, the longest wavelength corresponds to the transition from $$n = 3$$ to $$n = 2$$.

For $$\mathrm{Li}^{2+}$$ ($$Z = 3$$), the wavelength is scaled by $$\frac{1}{Z^2}$$.

So the ratio of wavelengths in hydrogen to that in $$\mathrm{Li}^{2+}$$:
$$\lambda_{Li^{2+}} = \frac{\lambda_{H}}{Z^2}$$

For the Balmer series, using the longest wavelength (which is 4 times the shortest wavelength in the Lyman series):
$$\lambda_{H(Balmer)} = 4 \lambda_{H(Lyman)} = 4x$$

Thus:
$$\lambda_{Li^{2+}} = \frac{4x}{3^2} = \frac{4x}{9}$$

$$\boxed{(D) \ \frac{4x}{9}}$$

Q14. de Broglie wavelength of a cricket ball of mass 0.5 kg moving with a speed of $$100 \mathrm{~ms}^{-1}$$ is-
(A) $$1.32 \times 10^{-34} \mathrm{~m}$$
(B) $$1.32 \times 10^{-35} \mathrm{~m}$$
(C) $$6.6 \times 10^{-28} \mathrm{~m}$$
(D) $$6.6 \times 10^{-34} \mathrm{~m}$$

(B) $$1.32 \times 10^{-35} \mathrm{~m}$$

The de Broglie wavelength of a cricket ball of mass $$0.5 \text{ kg}$$ moving with a speed of $$100 \text{ m/s}$$ is:

The de Broglie wavelength ($$\lambda$$) is given by:
$$\lambda = \frac{h}{mv}$$

Given:
$$h = 6.625 \times 10^{-34} \text{ J s}$$
$$m = 0.5 \text{ kg}$$
$$v = 100 \text{ m/s}$$

$$\lambda = \frac{6.625 \times 10^{-34}}{0.5 \times 100}$$
$$\lambda = \frac{6.625 \times 10^{-34}}{50}$$
$$\lambda = 1.325 \times 10^{-35} \text{ m}$$

$$\boxed{(B) \ 1.32 \times 10^{-35} \text{ m}}$$

Q15. If de Broglie wavelength of an electron revolving in the 4th Bohrorbit be $$\lambda$$, then the circumference of this orbit will be-
(A) $$\lambda$$
(B) $$2 \lambda$$
(C) $$3 \lambda$$
(D) $$4 \lambda$$

(D) $$4 \lambda$$

If de Broglie wavelength of an electron revolving in the 4th Bohr orbit be $$\lambda$$, then the circumference of this orbit will be:

For an electron in the $$n$$-th Bohr orbit, the circumference is given by:
$$2\pi r = n \lambda$$

For the 4th Bohr orbit ($$n = 4$$):
$$2\pi r = 4 \lambda$$

$$\boxed{(D) \ 4 \lambda}$$

Q16. When an electron drops from higher energy state $$(n)$$ to the ground state in an atom, the maximum number of spectral lines that can be obtained is-
(A) $$n-1$$
(B) $$n$$
(C) $$n(n-1)$$
(D) $$\frac{n(n-1)}{2}$$

(D) $$\frac{n(n-1)}{2}$$

When an electron drops from a higher energy state $$n$$ to the ground state in an atom, the maximum number of spectral lines that can be obtained is:

The maximum number of spectral lines that can be obtained when an electron transitions from a higher energy state $$n$$ to the ground state is given by the formula:

$$\frac{n(n-1)}{2}$$

This formula arises because each possible transition between two energy levels corresponds to a spectral line, and the total number of such transitions is the combination of $$n$$ levels taken 2 at a time.

$$\boxed{(D) \frac{n(n-1)}{2}}$$

Q17. Wave number corresponding to the spectral line having highest value is obtained from Rydberg equation $$\left(\bar{v}=R_{H} Z^{2}\left(\frac{1}{n_{j}^{2}}-\frac{1}{n_{i}^{2}}\right)\right)$$ by putting $$n_{i}$$ equal to-
(A) 1
(B) 2
(C) 0
(D) $$\infty$$

(D) $$\infty$$

Wave number corresponding to the spectral line having the highest value is obtained from the Rydberg equation $$\left(\bar{v}=R_{H} Z^{2}\left(\frac{1}{n_{j}^{2}}-\frac{1}{n_{i}^{2}}\right)\right)$$ by putting $$n_{i}$$ equal to:

The wave number ($$\bar{v}$$) is highest when the difference in energy levels is greatest. For the Rydberg equation:

$$\bar{v} = R_H Z^2 \left( \frac{1}{n_j^2} – \frac{1}{n_i^2} \right)$$

The highest value occurs when $$\frac{1}{n_i^2}$$ is minimized, which is when $$n_i$$ approaches infinity.

$$\boxed{(D) \ \infty}$$

Q18. The wavelength of any spectral line for an electronic transition in the spectrum is inversely proportional to-
(A) number of electrons undergoing transition
(B) nuclear charge of the atom
(C) the difference in the energy levels involved in the transition
(D) velocity of the electron undergoing transition

(C) the difference in the energy levels involved in the transition

The wavelength of any spectral line for an electronic transition in the spectrum is inversely proportional to:

The wavelength of a spectral line ($$\lambda$$) is inversely proportional to the difference in energy levels ($$\Delta E$$). According to the energy-wavelength relationship:

$$\Delta E = \frac{hc}{\lambda}$$

$$\boxed{(C) \ \text{the difference in the energy levels involved in the transition}}$$

Q19. Transition of an electron from $$n=3$$ level to $$n=2$$ level results in a spectral line in the-
(A) IR region
(B) visible region
(C) UV region
(D) far infrared region

(B) visible region

Transition of an electron from $$n=3$$ level to $$n=2$$ level results in a spectral line in the:

For hydrogen, transitions to the $$n=2$$ level are part of the Balmer series, which is in the visible region of the electromagnetic spectrum.

$$\boxed{(B) \ \text{visible region}}$$

Q20. The value of uncertainty calculated from the equation $$\Delta x . \Delta p=\frac{h}{4 \pi}$$ is-
(A) lowest value
(B) highest value
(C) exact value
(D) approximate value

(A) lowest value

The value of uncertainty calculated from the equation $$\Delta x \cdot \Delta p = \frac{h}{4\pi}$$ is:

The Heisenberg Uncertainty Principle states that the product of the uncertainties in position ($$\Delta x$$) and momentum ($$\Delta p$$) is at least $$\frac{h}{4\pi}$$.

$$\boxed{(A) \ \text{lowest value}}$$

Q21. In the ground state of $$\mathrm{M}^{2+}$$ ion, the last electron has the four quantum numbers: $$3,2,-2$$ and $$+\frac{1}{2}$$. The atomic number of M is-
(A) 23
(B) 24
(C) 25
(D) 26

(D) 26

In the ground state of $$M^{2+}$$ ion, the last electron has the four quantum numbers: $$3, 2, -2$$ and $$+\frac{1}{2}$$. The atomic number of $$M$$ is:

The quantum numbers $$n=3, l=2, m_l=-2, m_s=+\frac{1}{2}$$ correspond to a 3d orbital. This configuration indicates that the electron was removed from a 3d orbital. To find the atomic number:

– For $$M^{2+}$$, the configuration without the two lost electrons would be:
$$(\text{Ar}) 3d^6$$
– Therefore, $$M$$ has 26 electrons (since it lost 2 to become $$M^{2+}$$).
– So, the atomic number is 26.

$$\boxed{(D) \ 26}$$

Q22. The velocity of an electron with de Broglie wavelength 66 nm is $$\left(h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}\right)-$$
(A) $$1.84 \times 10^{-14} \mathrm{~ms}^{-1}$$
(B) $$5.4 \times 10^{3} \mathrm{~ms}^{-1}$$
(C) $$1.1 \times 10^{3} \mathrm{~ms}^{-1}$$
(D) $$1.1 \times 10^{7} \mathrm{~ms}^{-1}$$

(D) $$1.1 \times 10^{7} \mathrm{~ms}^{-1}$$

The velocity of an electron with de Broglie wavelength 66 nm is $$\left(h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}\right)- \ The de Broglie wavelength (\(\lambda$$) is related to the momentum ($$p$$) by:

$$\lambda = \frac{h}{p}$$

And momentum $$p$$ is related to mass ($$m$$) and velocity ($$v$$) by:

$$p = mv$$

Combining these, we get:

$$\lambda = \frac{h}{mv}$$
$$v = \frac{h}{m \lambda}$$

For an electron ($$m_e = 9.1 \times 10^{-31} \, \text{kg}$$) and $$\lambda = 66 \times 10^{-9} \, \text{m}$$:

$$v = \frac{6.6 \times 10^{-34} \, \text{kg m}^2/\text{s}}{9.1 \times 10^{-31} \, \text{kg} \times 66 \times 10^{-9} \, \text{m}}$$
$$v \approx 1.1 \times 10^7 \, \text{ms}^{-1}$$

$$\boxed{(D) \ 1.1 \times 10^7 \, \text{ms}^{-1}}$$

Q23. The line with highest wavelength in the Paschen series of hydrogen spectrum $$\left(R=109670 \mathrm{~cm}^{-1}\right)$$ –
(A) $$1215 \, Å$$
(B) $$9380 \, Å$$
(C) $$18760 \, Å$$
(D) $$5331 \, Å$$

(C) $$18760 \, Å$$

The line with highest wavelength in the Paschen series of hydrogen spectrum $$\left(R=109670 \, \text{cm}^{-1}\right)$$:

The Paschen series corresponds to transitions to the $$n=3$$ level. The highest wavelength (lowest energy transition) in this series is from $$n=4$$ to $$n=3$$.

Using the Rydberg formula:
$$\frac{1}{\lambda} = R \left( \frac{1}{3^2} – \frac{1}{4^2} \right)$$
$$\frac{1}{\lambda} = 109670 \left( \frac{1}{9} – \frac{1}{16} \right)$$
$$\frac{1}{\lambda} = 109670 \left( \frac{16 – 9}{144} \right)$$
$$\frac{1}{\lambda} = 109670 \left( \frac{7}{144} \right)$$
$$\lambda \approx 18760 \, \text{Å}$$

$$\boxed{(C) \ 18760 \, \text{Å}}$$

Q24. The wave number of 1st Balmer line in hydrogen spectrum is $$15200 \mathrm{~cm}^{-1}$$. The wave number of 1 st Balmer line in the spectrum of $$\mathrm{Li}^{2+}$$ ion will be-
(A) $$76000 \mathrm{~cm}^{-1}$$
(B) $$136800 \mathrm{~cm}^{-1}$$
(C) $$15200 \mathrm{~cm}^{-1}$$
(D) $$60800 \mathrm{~cm}^{-1}$$

(B) $$136800 \mathrm{~cm}^{-1}$$

The wave number of 1st Balmer line in hydrogen spectrum is $$15200 \, \text{cm}^{-1}$$. The wave number of 1st Balmer line in the spectrum of $$\text{Li}^{2+}$$ ion will be:

The wave number ($$\bar{v}$$) for $$\text{Li}^{2+}$$ ion can be found by multiplying the wave number for hydrogen by $$Z^2$$, where $$Z$$ is the atomic number of lithium (3).

$$\bar{v}_{\text{Li}^{2+}} = \bar{v}_{\text{H}} \times Z^2$$
$$\bar{v}_{\text{Li}^{2+}} = 15200 \, \text{cm}^{-1} \times 3^2$$
$$\bar{v}_{\text{Li}^{2+}} = 15200 \, \text{cm}^{-1} \times 9$$
$$\bar{v}_{\text{Li}^{2+}} = 136800 \, \text{cm}^{-1}$$

$$\boxed{(B) \ 136800 \, \text{cm}^{-1}}$$

Q25. In the ground state, the radius of H -atom is 0.53 A . So its radius in the 1st excited state is-
(A) $$0.13 \, \text{Å}$$
(B) $$1.06 \, \text{Å}$$
(C) $$4.77 \, \text{Å}$$
(D) $$2.12 \, \text{Å}$$

(D) $$2.12 \text{Å}$$

In the ground state, the radius of H-atom is 0.53 Å. So its radius in the 1st excited state is:

For hydrogen, the radius of the $$n$$-th orbit is given by:

$$r_n = n^2 \times r_1$$

For the first excited state ($$n=2$$):

$$r_2 = 2^2 \times 0.53 \, \text{Å}$$
$$r_2 = 4 \times 0.53 \, \text{Å}$$
$$r_2 = 2.12 \, \text{Å}$$

$$\boxed{(D) \ 2.12 \, \text{Å}}$$

Q26. $$X$$ and $$Y$$ are two isotones with mass numbers 70 and 72 respectively. If the atomic number of $$X$$ be 34 , the atomic number of $$Y$$ will be-
(A) 36
(B) 30
(C) 34
(D) 32

(A) 36

$$X$$ and $$Y$$ are two isotones with mass numbers 70 and 72 respectively. If the atomic number of $$X$$ be 34, the atomic number of $$Y$$ will be:

Isotones are atoms with the same number of neutrons.

For $$X$$ (with mass number 70 and atomic number 34):
$$\text{Number of neutrons in } X = 70 – 34 = 36$$

For $$Y$$ (

with the same number of neutrons):
$$\text{Number of neutrons in } Y = 36$$
$$\text{Atomic number of } Y = 72 – 36 = 36$$

$$\boxed{(A) \ 36}$$

Q27. In the $$(n+1)$$ th quantum shell, the values of azimuthal quantum number ‘ $$l$$ ‘ will be-
(A) 0 to $$n$$
(B) 0 to $$(n+1)$$
(C) 0 to $$(n-1)$$
(D) 1 to $$(n+1)$$

(A) 0 to $$n$$

In the $$(n+1)$$-th quantum shell, the values of azimuthal quantum number ‘ $$l$$ ‘ will be:

For a given principal quantum number $$n$$, the azimuthal quantum number $$l$$ can take values from 0 to $$n-1$$.

So, for the $$(n+1)$$-th quantum shell:
$$l = 0 \text{ to } n$$

$$\boxed{(A) \ 0 \text{ to } n}$$

Q28. Which of the following contains a pair of isosters-
(A) $$\mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2} \mathrm{O}$$
(B) $$\mathrm{HCl}_{1} \mathrm{~F}_{2}$$
(C) $$\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HNO}_{3}$$
(D) $$\mathrm{NH}_{3}, \mathrm{PH}_{3}$$

(A) $$\mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2} \mathrm{O}$$

Isosters are molecules or ions that have the same number of atoms and the same number of valence electrons.

$$\text{H}_2\text{O} \text{ and } \text{N}_2\text{O}$$

Both have 10 valence electrons and similar structure.

$$\boxed{(A) \ \text{H}_2\text{O}, \text{N}_2\text{O}}$$

Q29. Which of the following electronic transition in H -atom involves maximum amount of energy absorption-
(A) $$n=3$$ to $$n=4$$
(B) $$n=2$$ to $$n=5$$
(C) $$n=3$$ to $$n=6$$
(D) $$n=1$$ to $$n=2$$

(D) $$n=1$$ to $$n=2$$

The energy difference between levels is given by:

$$\Delta E = 13.6 \left( \frac{1}{n_f^2} – \frac{1}{n_i^2} \right) \text{eV}$$

The maximum energy absorption occurs for the largest difference between initial and final energy levels, which is:

$$n=1 \text{ to } n=2$$

$$\boxed{(D) \ n=1 \text{ to } n=2}$$

Q30. The correct sequence of paramagnetic character of the given metals is-
(A) $$\mathrm{Mn} > \mathrm{Cr} > \mathrm{Zn}$$
(B) $$\mathrm{Fe} > \mathrm{Zn} > \mathrm{Co}$$
(C) $$\mathrm{Cr} > \mathrm{Fe} > \mathrm{Zn}$$
(D) $$\mathrm{Hg} > \mathrm{Mn} > \mathrm{Fe}$$

(A) $$\mathrm{Mn} > \mathrm{Cr} > \mathrm{Zn}$$

Paramagnetic character is due to the presence of unpaired electrons. The paramagnetic character decreases with a decrease in the number of unpaired electrons.

$$\text{Mn} > \text{Cr} > \text{Zn}$$

Manganese (Mn) has the highest number of unpaired electrons, followed by Chromium (Cr), and Zinc (Zn) has none.

$$\boxed{(A) \ \text{Mn} > \text{Cr} > \text{Zn}}$$

Q31. Atomic spectra of which of the following is similar to that of $$\mathrm{He}^{+}$$-
(A) H
(B) $$\mathrm{Li}^{+}$$
(C) Na
(D) He

(B) $$\mathrm{Li}^{+}$$

The atomic spectrum of an ion is similar to another ion if both have the same number of electrons. $$\mathrm{He}^{+}$$ has only one electron (like a hydrogen atom).

– $$\mathrm{Li}^{2+}$$ has 3 protons and 1 electron (similar to $$\mathrm{He}^{+}$$).

$$\boxed{(B) \ \mathrm{Li}^{2+}}$$

Q32. The energy of the electron in the ground state of H -atom is -13.6 eV . Its energy in the 4 th quantum shell will be-
(A) $$-0.54 eV$$
(B) $$-5.40 eV$$
(C) $$-0.85 eV$$
(D) $$-2.77 eV$$

(C) $$-0.85 eV$$

The energy of the electron in the ground state of H-atom is -13.6 eV. Its energy in the 4th quantum shell will be:

The energy of an electron in the $$n$$-th quantum shell of a hydrogen atom is given by:

$$E_n = \frac{E_1}{n^2} = \frac{-13.6 \, \text{eV}}{4^2} = \frac{-13.6 \, \text{eV}}{16} = -0.85 \, \text{eV}$$

$$\boxed{(C) \ -0.85 \, \text{eV}}$$

Q33. If the radius of 1st Bohr orbit of H-atom is $$a_{0}$$, the radius of the 3 rd Bohr orbit will be-
(A) $$3 \times a_{0}$$
(B) $$6 \times a_{0}$$
(C) $$9 \times a_{0}$$
(D) $$a_{0} / 9$$

(C) $$9 \times a_{0}$$

If the radius of 1st Bohr orbit of H-atom is $$a_0$$, the radius of the 3rd Bohr orbit will be:

The radius of the $$n$$-th orbit in the Bohr model is given by:

$$r_n = n^2 \times r_1$$

For the 3rd Bohr orbit ($$n = 3$$):

$$r_3 = 3^2 \times a_0 = 9 \times a_0$$

$$\boxed{(C) \ 9 \times a_0}$$

Q34. Which pair of ions have the same electronic configuration-
(A) $$\mathrm{Cr}^{3+}, \mathrm{Fe}^{3+}$$
(B) $$\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}$$
(C) $$\mathrm{Fe}^{3+}, \mathrm{Co}^{3+}$$
(D) $$\mathrm{Sr}^{3+}, \mathrm{Cr}^{3+}$$

(B) $$\mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}$$

$$\mathrm{Fe}^{3+}$$ and $$\mathrm{Mn}^{2+}$$

$$\mathrm{Fe}^{3+} : (\text{Ar}) \, 3d^5$$
$$\mathrm{Mn}^{2+} : (\text{Ar}) \, 3d^5$$

$$\boxed{(B) \ \mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}}$$

Q35. The radius of which of the following is equal to that of the 1st Bohr orbit of H -atom-
(A) $$\mathrm{He}^{+}(n=2)$$
(B) $$\mathrm{Li}^{2+}(n=2)$$
(C) $$\mathrm{Li}^{2}(n=3)$$
(D) $$\operatorname{Be}^{3+}(n=2)$$

(D) $$\operatorname{Be}^{3+}(n=2)$$

For $$\mathrm{Li}^{2+}$$ in the $$n = 2$$ state:

$$r_n = \frac{n^2 \times a_0}{Z} = \frac{2^2 \times a_0}{3} = \frac{4a_0}{3}$$

For $$\mathrm{He}^{+}$$ in the $$n = 2$$ state:

$$r_n = \frac{2^2 \times a_0}{2} = 2a_0$$

For $$\mathrm{Be}^{3+}$$ in the $$n = 2$$ state:

$$r_n = \frac{2^2 \times a_0}{4} = a_0$$

$$\boxed{(D) \ \mathrm{Be}^{3+} (n=2)}$$

Q36. Electromagnetic radiation with least wavelength is-
(A) ultraviolet ray
(B) cosmic ray
(C) X-ray
(D) infrared ray

(B) cosmic ray

The electromagnetic spectrum is ordered from longest to shortest wavelength as follows: infrared < visible < ultraviolet < X-rays < gamma rays < cosmic rays.

$$\boxed{(B) \ \text{cosmic ray}}$$

Q37. wavelength of photon with energy $$3.03 \times 10^{-19} \mathrm{~J}$$ is-
(A) 65.6 nm
(B) 656 mm
(C) 0.656 nm
(D) 6.56 nm

(A) 65.6 nm

$$E = \frac{hc}{\lambda}$$

$$\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{3.03 \times 10^{-19} \, \text{J}}$$

$$\lambda \approx 656 \, \text{nm}$$

$$\boxed{(A) \ 656 \, \text{nm}}$$

Q38. Correct sequence of number of unpaired electrons of the given ions is-
(A) $$\mathrm{Cu}^{2+} > \mathrm{Ni}^{2+} > \mathrm{Cr}^{3+} > \mathrm{Fe}^{3+}$$
(B) $$\mathrm{Fe}^{3+} > \mathrm{Cr}^{3+} > \mathrm{Ni}^{2+} > \mathrm{Cu}^{2+}$$
(C) $$\mathrm{Ni}^{2+} > \mathrm{Cu}^{2+} > \mathrm{Fe}^{3+} > \mathrm{Cr}^{3+}$$
(D) $$\mathrm{Cr}^{3+} > \mathrm{Fe}^{3+} > \mathrm{Cu}^{2+} > \mathrm{Ni}^{2+}$$

(B) $$\mathrm{Fe}^{3+} > \mathrm{Cr}^{3+} > \mathrm{Ni}^{2+} > \mathrm{Cu}^{2+}$$

$$\mathrm{Cu}^{2+} : (\text{Ar}) \, 3d^9$$
$$\mathrm{Ni}^{2+} : (\text{Ar}) \, 3d^8$$
$$\mathrm{Cr}^{3+} : (\text{Ar}) \, 3d^3$$
$$\mathrm{Fe}^{3+} : (\text{Ar}) \, 3d^5$$

The correct sequence is:

$$\boxed{(B) \ \mathrm{Fe}^{3+} > \mathrm{Cr}^{3+} > \mathrm{Ni}^{2+} > \mathrm{Cu}^{2+}}$$

Q39. If the energy of 2 nd Bohr orbit be $$-328 \mathrm{kJmol}^{-1}$$, the energy of the 4 th Bohr orbit will be-
(A) $$-41 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
(B) $$-1312 \mathrm{kJmol}^{-1}$$
(C) $$-164 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
(D) $$-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

(D) $$-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

Energy levels are given by:

$$E_n = \frac{E_1}{n^2}$$

Given $$E_2 = -328 \, \text{kJ/mol}$$:

$$E_1 = -328 \times 4 = -1312 \, \text{kJ/mol}$$

For $$n=4$$:

$$E_4 = \frac{-1312}{4^2} = \frac{-1312}{16} = -82 \, \text{kJ/mol}$$

$$\boxed{(D) \ -82 \, \text{kJ/mol}}$$

Q40. Which of the following $$d$$-orbitals does not contain four lobes-
(A) $$d_{x^{2}-y^{2}}$$
(B) $$d_{x y}$$
(C) $$d_{y z}$$
(D) $$d_{z^{2}}$$

(D) $$d_{z^{2}}$$

$$d_{z^2}$$ has a unique shape with two lobes along the z-axis and a torus around the middle.

$$\boxed{(D) \ d_{z^2}}$$

Q41. The wave number of the 1st Balmer line in the hydrogen spectrum ( $$R=$$ Rydberg constant) is-
(A) $$\frac{9 R}{400} \mathrm{~cm}^{-1}$$
(B) $$\frac{7 R}{144} \mathrm{~cm}^{-1}$$
(C) $$\frac{3 R}{4} \mathrm{~cm}^{-1}$$
(D) $$\frac{5 R}{36} \mathrm{~cm}^{-1}$$

(D) $$\frac{5 R}{36} \mathrm{~cm}^{-1}$$

For the first Balmer line ($$n=3$$ to $$n=2$$):

$$\bar{\nu} = R \left( \frac{1}{2^2} – \frac{1}{3^2} \right)$$

$$\bar{\nu} = R \left( \frac{1}{4} – \frac{1}{9} \right)$$

$$\bar{\nu} = R \left( \frac{5}{36} \right)$$

$$\boxed{(D) \ \frac{5R}{36} \, \text{cm}^{-1}}$$

Q42. Number of nodal planes of $$f$$-orbital is-
(A) 1
(B) 2
(C) 3
(D) 4

(C) 3

An $$f$$-orbital has 3 nodal planes.

$$\boxed{(C) \ 3}$$

Q43. Kinetic energy of the electron in the 2nd Bohr orbit of H -atom ( $$a_{0}=$$ Bohr radius of H -atom) is-
(A) $$\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}$$
(B) $$\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$$
(C) $$\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$$
(D) $$\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}$$

(B) $$\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$$

The total energy $$E$$ in the $$n$$-th orbit is:

$$E_n = -\frac{13.6 \, \text{eV}}{n^2}$$

The kinetic energy $$K$$ is half the magnitude of the total energy but positive:

$$K = -\frac{E_n}{2}$$

For $$n=2$$:

$$E_2 = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV}$$
$$K = -\frac{-3.4 \, \text{eV}}{2} = 1.7 \, \text{eV}$$

In terms of the given options, using the expression for kinetic energy:

$$K = \frac{h^2}{16 \pi^2 m a_0^2}$$

$$\boxed{(B) \ \frac{h^2}{16 \pi^2 m a_0^2}}$$

Q44. The maximum number of electrons in an atom with the quantum numbers $$n=3$$ and $$m_{s}=-\frac{1}{2}$$ is-
(A) 4
(B) 7
(C) 9
(D) 16

(C) 9

For $$n=3$$:

– $$l$$ can be 0, 1, or 2.
– For each $$l$$, $$m_l$$ can have $$(2l + 1)$$ values.
– Each $$m_l$$ can have 2 $$m_s$$ values ($$+\frac{1}{2}$$ and $$-\frac{1}{2}$$).

The

maximum number of electrons with $$m_s = -\frac{1}{2}$$ is half the total number of electrons in $$n=3$$:

$$2n^2 = 2 \times 3^2 = 18$$

Half of these have $$m_s = -\frac{1}{2}$$:

$$18/2 = 9$$

$$\boxed{(C) \ 9}$$

Q45. In Sommerfeld’s modification of Bohr’s theory, the trajectory of an electron in a hydrogen atom is-
(A) perfect ellipse
(B) a closed ellipse like curve, narrower at the perihelion position and flatter at the aphelion position
(C) a closed loop on spherical surface
(D) a rosette

(A) perfect ellipse

In Sommerfeld’s model, the orbit of an electron is an ellipse.

$$\boxed{(A) \ \text{perfect ellipse}}$$

Q46. Identify the species (with its charge) having 16 protons, 18 electrons and 16 neutrons-
(A) $$\mathrm{S}^{-}$$
(B) $$\mathrm{Si}^{2-}$$
(C) $$\mathrm{P}^{3-}$$
(D) $$\mathrm{S}^{2-}$$

(D) $$\mathrm{S}^{2-}$$

Protons = 16 (so element is Sulfur, S)
Electrons = 18 (so the ion has 2 extra electrons, $$S^{2-}$$)

$$\boxed{(D) \ \mathrm{S}^{2-}}$$

Q47. Probability of finding the electron in $$y z$$-plane is zero for the orbital-
(A) $$p_{x}$$
(B) $$p_{y}$$
(C) $$p_{z}$$
(D) $$d_{y z}$$

(A) $$p_{x}$$

The $$p_x$$ orbital is oriented along the x-axis and has zero probability in the $$yz$$-plane.

$$\boxed{(A) \ p_x}$$

Q48. Energy required for the fission of 1 mol of $$\mathrm{Cl}-\mathrm{Cl}$$ bond in $$\mathrm{Cl}_{2}$$ is $$242 \mathrm{~kJ} \mathrm{~mol}^{-1}$$. Thus the wavelength of the light required to break $$\mathrm{Cl}-\mathrm{Cl}$$ bond is (given: $$c=3 \times 10^{8} \mathrm{~ms}^{-1}$$ and $$N_{A}=6.02 \times 10^{23} \mathrm{~mol}^{-1} \mid$$ –
(A) 594 nm
(B) 640 nm
(C) 700 nm
(D) 494 nm

(D) 494 nm

$$E = 242 \times 10^3 \, \text{J/mol}$$
$$E_{\text{photon}} = \frac{242 \times 10^3 \, \text{J/mol}}{6.02 \times 10^{23} \, \text{mol}^{-1}} \approx 4.02 \times 10^{-19} \, \text{J}$$

$$\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{4.02 \times 10^{-19} \, \text{J}} \approx 494 \, \text{nm}$$

$$\boxed{(D) \ 494 \, \text{nm}}$$

Q49. An electron is revolving at a speed of $$600 \mathrm{~ms}^{-1}$$ (accuracy $$=0.005 \%$$ ). The uncertainty in the determination of its position is (given: $$h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} 5^{-1}, e_{m}=9.1 \times 10^{-31} \mathrm{~kg}$$ l-
(A) $$1.92 \times 10^{-3} \mathrm{~m}$$
(B) $$3.84 \times 10^{-3} \mathrm{~m}$$
(C) $$1.52 \times 10^{-4} \mathrm{~m}$$
(D) $$5.10 \times 10^{-3} \mathrm{~m}$$

(A) $$1.92 \times 10^{-3} \mathrm{~m}$$

$$\Delta v = 600 \times 0.005\% = 600 \times 0.00005 = 0.03 \, \text{m/s}$$

$$\Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.63 \times 10^{-34}}{4\pi (9.1 \times 10^{-31})(0.03)} \approx 1.92 \times 10^{-3} \, \text{m}$$

$$\boxed{(A) \ 1.92 \times 10^{-3} \, \text{m}}$$

Q50. Which of the following ions has the electronic configuration $$(\mathrm{Ar}) 3 d^{6}-$$
(A) $$\mathrm{Co}^{3+}$$
(B) $$\mathrm{Ni}^{3+}$$
(C) $$\mathrm{Mn}^{3+}$$
(C) $$\mathrm{Fe}^{3+}$$

(A) $$\mathrm{Co}^{3+}$$

To find the electronic configuration of each ion, we start with the neutral atom and then remove electrons corresponding to the charge of the ion:

1. $$\mathrm{Co}^{3+}$$:
– Neutral Co: $$(\mathrm{Ar}) 3d^7 4s^2$$
– Remove 3 electrons: $$(\mathrm{Ar}) 3d^6$$
– Configuration: $$(\mathrm{Ar}) 3d^6$$

2. $$\mathrm{Ni}^{3+}$$:
– Neutral Ni: $$(\mathrm{Ar}) 3d^8 4s^2$$
– Remove 3 electrons: $$(\mathrm{Ar}) 3d^7$$
– Configuration: $$(\mathrm{Ar}) 3d^7$$

3. $$\mathrm{Mn}^{3+}$$:
– Neutral Mn: $$(\mathrm{Ar}) 3d^5 4s^2$$
– Remove 3 electrons: $$(\mathrm{Ar}) 3d^4$$
– Configuration: $$(\mathrm{Ar}) 3d^4$$

4. $$\mathrm{Fe}^{3+}$$:
– Neutral Fe: $$(\mathrm{Ar}) 3d^6 4s^2$$
– Remove 3 electrons: $$(\mathrm{Ar}) 3d^5$$
– Configuration: $$(\mathrm{Ar}) 3d^5$$

$$\boxed{(A) \ \mathrm{Co}^{3+}}$$

Thank You

#### Major Branches of Biology

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