NEET Practice Questions (MCQs) with Answers & Solutions

(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.

(a) n represents the main energy level and l represents the subshell.

lf n = 4 and l = 3, the subshell is 4f.

In f subshell, there are 7 orbitals and each orbital can accommodate a maximum of two electrons, so, maximum number of elecirons in 4f subshell = 7x 2 = 14

Number of atomic orbitals in an orbit = n² = 4² =16

(a)

 E₁ = 25eV, E₂ = 50eV

$$\begin{gathered} {\mathrm{E}}_1 = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_1} \mathrm{and} {\mathrm{E}}_2 =\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2} \\ \mathrm{or} \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1} \\ \mathrm{or} \frac{25}{50}=\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1} \\ \mathrm{or} {\mathrm{\lambda }}_1 = 2{\mathrm{\lambda }}_2 \end{gathered}$$

(d) Key Idea Write the electronic configurations of given ions and find the correct answer.

           Ni³⁺ (28) = [Ar]3d⁷

           Mn³⁺ (25) = [Ar]3d⁴

           Fe³⁺ (26) = [Ar]3d⁵

           Co³⁺ (27) = [Ar]3d⁶

Key Idea Number of oxidation states exhibited by d-block elements is the sum of number of electrons (unpaired) in d-orbitals and number of electrons in s-orbital.

(a) 3d³, 4s² $\Rightarrow$ O.S = 3 + 2 = 5

(b) 3d⁵, 4s¹ $\Rightarrow$ O.S = 5 + 1 = 6

(c) 3d⁵, 4s² $\Rightarrow$ O.S = 5 + 2 = 7

(d) 3d², 4s² $\Rightarrow$ O.S = 2 + 2 = 4

Hence, element with 3d⁵, 4s² configuration exhibits largest number of oxidation states.

Total number of subshells = (2l+1)

$\therefore$Maximum number of electrons m the subshell = 2(2l+1)= 4l+2