Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature but somewhat higher pressure).
I2(s) → I2(g)
I2(s) → I2(l)
Is ΔS positive or negative in these processes?

ΔS increases in each case.  

Explanation:

Entropy (S) is a measure of the degree of the spreading of energy within a system. The more ways that energy can be distributed, the greater the entropy.

(a) Sublimation of I₂

I₂(s)⇌ I₂(g)

In the solid, the I₂ molecules are locked in position in a crystal lattice. They can vibrate only slightly about their equilibrium locations.  

Those in the gas phase can move in different directions and at different speeds, so the thermal energy is widely distributed among many microstates. The mobility of the molecules is greatly increased, so the number of number of microstates increases and entropy increases.

(b) Melting of I₂

I₂(s)⇌ I₂(ℓ)

The molecules in the solid can only vibrate about a fixed position.

Those in the liquid phase can move around and slide past each other. The increased freedom of motion means that thermal energy is distributed among more microstates, so the entropy increases.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

the answer is d. 74.02