Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in $z$-direction] ?

A. $2 \mathrm{p}_{\mathrm{z}}$ and $2 \mathrm{p}_{\mathrm{x}}$

B. 2 s and $2 \mathrm{p}_{\mathrm{x}}$

C. $3 d_{x y}$ and $3 d_{x^2-y^2}$

D. 2 s and $2 \mathrm{p}_{\mathrm{z}}$

E. $2 p_z$ and $3 d_{x^2-y^2}$

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Which of the following statement is true with respect to $\mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{CH}_4$ ?

A. The central atoms of all the molecules are $\mathrm{sp}^3$ hybridized.

B. The $\mathrm{H}-\mathrm{O}-\mathrm{H}, \mathrm{H}-\mathrm{N}-\mathrm{H}$ and $\mathrm{H}-\mathrm{C}-\mathrm{H}$ angles in the above molecules are $104.5^{\circ}, 107.5^{\circ}$ and $109.5^{\circ}$, respectively.

C. The increasing order of dipole moment is $\mathrm{CH}_4<\mathrm{NH}_3<\mathrm{H}_2 \mathrm{O}$.

D. Both $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ are Lewis acids and $\mathrm{CH}_4$ is a Lewis base.

E. A solution of $\mathrm{NH}_3$ in $\mathrm{H}_2 \mathrm{O}$ is basic. In this solution $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ act as Lowry-Bronsted acid and base respectively.

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Match the List - I with List - II

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Arrange the following compounds in increasing order of their dipole moment :

$\mathrm{HBr}, \mathrm{H}_2 \mathrm{~S}, \mathrm{NF}_3$ and $\mathrm{CHCl}_3$