The structure of H₂SO₄ is 

Total number of unshared electrons = 10 x 2 = 20

Therefore, among all the given options, option a is correct.

In modern Periodic Table, the physical and chemical properties of the elements are a periodic function of their atomic number, i.e., if the elements are arranged in order of their increasing atomic number, the elements with similar properties are repeated after certain regular intervals.

The quantum numbers + $\frac{1}{2}$ and - $\frac{1}{2}$ indicate the rotation of electrons around its axis in opposite directions. Hence, they represent the two quantum mechanical spin states which have no classical analogues.

Given, p = 745 mm = $\frac{745}{760} \mathrm{atm}$

T = 65$°$C = 65 + 273 = 338 K

$$\begin{gathered} \therefore \mathrm{M} = \frac{\mathrm{dRT}}{\mathrm{p}} = \frac{1.56 \times 0.0821 \times 338}{{\displaystyle \frac{745}{760}}} \\ = 44.2 \mathrm{u} \end{gathered}$$

CH₃ - C $\equiv$ C - H has a bond formed by overlap of sp-sp³ hybrid orbitals. Other options consist of following orbitals:

b. CH₃ - CH = CH - CH₃ 

It has a bond overlapped by sp³-sp²-sp²-sp³ orbital.

c. CH₂ = CH - CH = CH₂

It has a bond overlapped by sp² orbital.

d. HC $\equiv$CH 

It has a bond overlapped by sp orbital.

The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation, is called its ionisation energy. Because the size of C is smaller than that of the B due to effective nuclear charge, hence

I^st IE of C > I^st IE of B

The ionisation energy of 1s², 2s², 2p³ configuration is highest due to presence of half- filled orbitals.

N₂( 7 + 7 = 14e^-) = KK^*, $\mathrm{\sigma }$2s², $\stackrel{*}{\mathrm{\sigma }}$2s², $\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2$

                           $\approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2 , \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

                   BO    = $\frac{1}{2}$(8 - 2) = 3

O₂ (8 + 8 =16e^-) = KK^*, $\mathrm{\sigma }$2s², $\stackrel{*}{\mathrm{\sigma }}$2s², $\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

                             $\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2 , \stackrel{*}{\mathrm{\pi }} 2{\mathrm{p}}_{\mathrm{x}}^1 \approx \stackrel{*}{\mathrm{\pi }} 2{\mathrm{p}}_{\mathrm{y}}^1$

                   BO   = $\frac{1}{2} (8 - 4) = 2$

He₂ ( 2 + 2 = 4e^-) = $\sigma$1s² , $\mathrm{\sigma }$^*1s²

                       BO = $\frac{1}{2}$(2 - 2) = 0

H₂ (1 + 1 = 2e^-) = $\mathrm{\sigma }$1s²

                  BO   = $\frac{1}{2}$(2 - 0) = 1

Hence, N₂ has the highest bond order.

Mn in [Mn(H₂O)₆]²⁺ is present as Mn²⁺(3d⁵).

sp³d² hybridisation occurs in[Mn(H₂O)₆]⁺². Due to the presence of weak field ligand pairing of electron does not take place, hence the number of unpaired electrons is five.

Bond angle is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a  molecule/complex ion.

s character $\propto$ bond angle

Hence, the correct order is sp³ < sp² < sp