## The molecule which possesses both sp3 and sp3d2 hybridisation is

Q: The molecule which possesses both sp^{3} and sp^{3}d^{2} hybridisation is

(A) Solid PCl_{5}

(B) Gaseous PCl_{5}

(C) PCl_{4}

(D) PCl_{6}

Ans: (A)

Atomic Structure

Q: The molecule which possesses both sp^{3} and sp^{3}d^{2} hybridisation is

(A) Solid PCl_{5}

(B) Gaseous PCl_{5}

(C) PCl_{4}

(D) PCl_{6}

Ans: (A)

Question: Two particles A and B are in motion. If the wavelength associated with particle A is 5 × 10^{–8} m, calculate the wavelength associated with particle B if its momentum is half of A.

Solution: According to de-Broglie equation

$\lambda_A = \frac{h}{p_A}$ and $\lambda_B = \frac{h}{p_B}$

$\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A}$

But p_{B} = (1/2) p_{A} (given)

$\frac{\lambda_A}{\lambda_B} = \frac{1}{2}$

λ_{B} = 2 λ_{A}

λ_{B} = 2 × 5 × 10^{–8} m

= 10^{–7} m

Q: Why electron cannot exist inside the nucleus according to Heisenberg’s uncertainty principle ?

Sol: Diameter of the atomic nucleus is of the order of 10^{–15} m

The maximum uncertainty in the position of electron is 10^{–15} m

Mass of electron = 9.1 × 10^{–31} kg

$\large \Delta x . \Delta p = \frac{h}{4\pi}$

$\large \Delta x . ( m \Delta v ) = \frac{h}{4\pi}$

$\large \Delta v = \frac{h}{4\pi} . \frac{1}{\Delta x m}$

$\large \Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.14} . \frac{1}{10^{-15} \times 9.1 \times 10^{-31}}$

Δv = 5.80 × 10^{10} ms^{–1}

This value is much higher than the velocity of light and hence not possible.

Q: Which of the following is the nodal plane of d_{xy} orbital ?

(A) XY

(B) YZ

(C) ZX

(D) all

Ans: (B) & (C)

Q: Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln(An/A1) against ln(n):

(A) will pass through origin

(B) will be straight line with slope = 4

(C) will be a monotonically increasing non-linear curve

(D) will be a circle

Ans: (A) , (B) & (C)