Chemistry MCQs for NEET — Practice Questions with Answers

(a) Magnetic moment, $\mathrm{\mu } = \sqrt{\mathrm{n}(\mathrm{n}+2)}$ BM where,

n = number of unpaired electrons
                         μ = 2.84 (given)
$\therefore$                  284 = $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ BM
                  (2.84)²= n(n + 2)
                          8 = n² + 2n
             n² + 2n -8 = 0
      n² + 4n - 2n -8 = 0

                          n = 2

$$\begin{gathered} {\mathrm{Ni}}^{2+} = \left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^0 (\mathrm{two} \mathrm{unpaired} \mathrm{electrons}) \\ {\mathrm{Ti}}^{3+} = \left[\mathrm{Ar}\right]3{\mathrm{d}}^{1}4{\mathrm{s}}^0 (\mathrm{one} \mathrm{unpaired} \mathrm{electrons}) \\ {\mathrm{Cr}}^{3+} =\left[\mathrm{Ar}\right]3{\mathrm{d}}^3 (\mathrm{three} \mathrm{unpaired} \mathrm{electrons}) \\ {\mathrm{Co}}^{2+}= \left[\mathrm{Ar}\right]3{\mathrm{d}}^7,4{\mathrm{s}}^0 (\mathrm{three} \mathrm{unpaired} \mathrm{electrons}) \\ \mathrm{So}, \mathrm{only} {\mathrm{Ni}}^{2+} \mathrm{has} 2\mathrm{unpaired} \mathrm{electrons}. \end{gathered}$$

Electronic configuration of Fe²⁺ is [Ar]3d⁶4s⁰.

.^.. Number of electrons = 6

                Mg - 1s²2s²2p⁶3s² (6s electrons)

It matches with the 6d electrons of Fe²⁺

                Cl - 1s²2s²2p⁶3s²3p⁵ (11p electrons)

It does not match with the 6d electrons of Fe²⁺

                Ne-1s²2s²2p⁶ (6p electrons)

It matches with the 6d electrons of Fe²⁺.

Hence, Cl has 11p electrons which do not match in number with 6d electrons of Fe²⁺.

(a) Angular momentum of electron in d-orbital is

              $$\begin{gathered} =\sqrt{\mathrm{l}(\mathrm{l}+1)}\frac{\mathrm{h}}{2\mathrm{\pi }}; \mathrm{for} \mathrm{d}-\mathrm{orbital}, \mathrm{l}=2 \\ =\sqrt{2(2+!)} \Rightarrow \mathrm{h}=\sqrt{6} \mathrm{h} \left(\because \mathrm{h}=\frac{\mathrm{h}}{2\mathrm{\pi }}\right) \end{gathered}$$

d.

The correct order of increasing energy of the listed orbitals in the atom of titanium is $3s<3p<4s<3d$.

The energy of 4s orbital is lower than the energy of 3d orbital.

Due to this, 4s orbital is filled first followed by 3d orbital.

The wavelength of light is related to its energy by the equation,

 E=hc/$\mathrm{\lambda }$.

Given, $\mathrm{\lambda }$ = 45 nm = 45 x 10^-9 m [._.. 1 nm = 10^-9 m]

Hence, E = (6.63 x 10^-34 Js x 3 x 10⁸ ms-1)/45 x 10^-9 m = 4.42 x 10^-18 J

Hence, the energy corresponds to light of wavelength 45 nm is 4.42 x 10^-18 J.

(c) Given, Planck's constant,

            h = 6.63 x 10^-34 J-s

Speed of light, c = 3 x 10¹⁷ nms^-1

Frequency of quanta

              v = 6 x 10¹⁵ s^-1

 Wavelength, $\mathrm{\lambda }$ = ?

We know that, v = c/$\mathrm{\lambda }$

(d) The orbital of the electron having n=3, l=1 and m=-1 is 3p_z (as nl_m) and an orbital can have a maximum of two electrons with opposite spins.

$\therefore$ 3p_z orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1.

(d) 

$$\begin{gathered} \mathrm{If} \mathrm{n}=1, \\ {\mathrm{E}}_1=-2.178 \times {10}^{-18}{\mathrm{Z}}^2\mathrm{J} \\ \mathrm{If} \mathrm{n}=6 \\ {\mathrm{E}}_6=\frac{-2.178 \times {10}^{-18} {\mathrm{Z}}^2}{36}\mathrm{J} \\ = 6.05 \times {10}^{-20}{\mathrm{Z}}^2\mathrm{J} \end{gathered}$$

From the above calculation, it is obvious that electron has a more negative energy than it does for n 6. It means that electron is more strongly bound in the smallest allowed orbit.