H₂S-In the central atom is sulphur. There are 6 electrons in its valence shell (₁₆S = 2,8,6).Two electrons are shared with two H-atoms and the remaining four electrons are present as two one pair. Hence, total pairs of electrons are four(2 bond pairs and 2 lone pairs).
Due to the presence of 2 lone pairs the shape becomes distorted tetrahedral or angular or bent (non-linear).
PCl₃-In the central atom is phosphorus. There are 5 electrons in its valence shell (₁₆P = 2,8,5).Three electrons are shared with three Cl-atoms and the remaining two electrons are present as one long pair. Hence, total pairs of electrons are four(1 lone pairs and 3 bond pairs). Due to the presence of one lone pair, the shape becomes pyramidal (non-planar).

A covalent bond is formed by the overlapping of atomic orbitals. The direction of overlapping gives the direction of bond. In ionic bond, the electrostatic field of an ion is non-directional. Each positive ion is surrounded by a number of anions in any direction depending upon its size and vice-versa.
That's why covalent bonds are directional bonds while ionic bonds are non-directional.

In ethyne molecule, both the carbon atoms are sp hybridised having two unhybridised orbitals, i.e. 2P_x and 2P_y. The two sp hybrid orbitals of both the carbon atoms are oriented in opposite direction forming an angle of 180^o.
\(H-C\overset { \sigma }{ \underset { 2\pi -bond }{ \equiv } C } -H\)
That's why ethyne molecule is linear.

Z has seven electrons in its valence shell. It is the most electronegative element. Therefore, HZ will have the highest dipole moment.

Formal charge on an atom in a Lewis structure
= [total number of valence electrons in free atom]
- [total number of non-bonding (lone pairs) electrons]
- \(\frac { 1 }{ 2 } \)[total number of bonding or shared electron]
Formal charge on H = 1 - 0 x 2 = 0
Formal charge on N = 5 - 0 x 8 = 1
Formal charge on O (1) = 6 - 4 - x 4 = 0
Formal charge on O (2) = 6 - 4 - x 4 = 0
Formal charge on O (3) = 6 - 6 - x 2 = -1


On solving we get, formal charges on O(1), N and O(2) as 0, +1 and -1 respectively. 
Similarly, on solving we get formal charge on H(1), H(2), O(2), O(3), O(4) ans S, as 0, 0, 0, -1 , 0, -1 and +2 respectively.