NEET Practice Questions (MCQs) with Answers & Solutions

(b) The element with atomic number, Z = 114 is flerovium (Fl). It is a super heavy artificial chemical element. In the periodic table of the elements, it is transactinide element in the p-block. It is a member of 7th period and is the heaviest known member of the carbon family.

Electronic configuration for Z =114 is

        [Rn]₈₆5f¹⁴,6d¹⁰,7s²,7p²

(b)

We know that n=3, l=1 is for 3p-orbital,so maximum 6 electrons can be filled.

The d_z² and d_x²-y² orbitals have electron density along the x, y, and z axes. The d_z² orbital has a dumbbell shape with electron density along the z-axis, while the d_x²-y² orbital has a double dumbbell shape with electron density along the x and y axes.

(c) Two electrons occupying the same orbital has equal spin but the directions of their spin are opposite. Hence, spin quantum number s (represented +1/2 and -1/2) distinguishes them.

(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.