Test: VSEPR Theory & Hybridisation - NEET MCQ

In each central atom is sp²-hybridised, each having one lone pair.
Difference in the extent of repulsion changes value of  θ.

 XeF₂-sp³d hybridised Xe
Three lone pairs on equatorial positions to minimise repulsion; two F-atom at axial position. Thus, it is linear.

(a) TeF₂,Te₅₀ - Six electrons in valence shell

(b)   hybridised l- atom with three lone pairs at equatorial positions to minimise repulsion thus, linear.

(c) SoCI₃ - sp³-hybridised pyramidal Sb-one lone pair on Sb.
(d) BaCI₂ - ionic .

(b) BrF₃ - T-shaped (sp³d²)

Both have one lone pair-trigonal pyramidal. 

Has no lone pair thus, bond angle is 120°. B-atom is sp²-hybridised.

Central atom E is sp³-hybridised. Hence, angle < 120°. As we go down the group, (Ip-bp) repulsion decreases. Hence, angle 
(Cl—E—Cl) PCI₃ > AsCI₃ > BiCI₃.
Thus, order is BCI₃ > PCI₃ > AsCI₃ > BiCI₃.

1 has three lone pair and two bond pair, its representation will be AB2L3 and according to VSEPR theory it is linear in shape.

2 has two LP and two BP which representation will be AB2L2, so according to VSEPR theory it will be bent or it can also said to be 'V' shaped.

Hence B is correct.

PCI₅
sp³d-hybrid orbitals—No non-bonding pair trigonal bipyramidal structure.

IF₅ sp³d²-hybrid orbitals—One non-bonding pair; one F and non-bonding pair lie in different sides so as to minimise repulsion. Square pyramidal.

There is (i) Ip - Ip (ii) Ip - bp (iii) bp - bp repulsion
Thus, bond angle is less than 109.5°. In H₂O, 104.5° > 90° in H₂S. It is due to larger size of S-atom which minimises the repulsion and allows the bonds in H₂S to be purely p-type.

The high electronegativity of F pulls the bonding electrons farther away from N than in NH₃. Thus, repulsion between bond pairs is iess in NF₃ than in NH₃. Flence, the lone pair in N causes a greater distortion than NH₃.
(b) as in (a)

Going down the group in periodic table, as size of central atom increases, repulsion increases.

O-atorn is sp³-hybridised and thus, bent due to presence of lone pairs on O-atom molecule is polar and planar.

Hence A and B only

NH₃   N atom sp³ -hybridised   - Bipyramidal
H₂O   O a to m sp³ -hybridised   - V -shaped
H²S   S atom is sp³ -hybridised   - Bent
SO₂   S atom is sp² -hybridised   - Bent
Ni(CO)₄   Ni atom is sp³-hybridised   - Tetrahedral
[NiCI4]^2-  Ni atom is sp³-hybridised   -Tetrahedral.
SF₄   S atom is sp³d -hybridised   - See-saw
PCI⁵   P atom is sp³ -hybridised   - Bipyramidal

Hybridisation of l-True
(b) Oxidation state of I = + 7 (highest) -True.
(c) It has two types of bonds two axial (shortest) and five equatorial (longest) - True.

(d) Pentagonal bipyramidai structure - True.

 sp³d -hybridised Cl-atom - (T-shaped s tructure)
  sp³ -hybridised. Cl-atom (Bent)
 sp³d -hybridised As-atom (Trigonal bipyramidai).
 sp³d² -hybridised As-atom (Octahedral).

 sp³d -hybridised Cl-atom - (T-shaped s tructure)
  sp³ -hybridised. Cl-atom (Bent)
 sp³d -hybridised As-atom (Trigonal bipyramidai).
 sp³d² -hybridised As-atom (Octahedral).

Based on VSEPR model, it is (T-shaped).
Thus, (i) → (r)


AX₄E₂ (Square planar)

Thus, (ii) → (s)
(iii)   lone pair = 1, AX₄E (See-saw)

Thus, (iii) → (q) 
lone pair = 3

Linear structure
Three lone pairs are at equatorial position to minimise repulsion.
Linear structure.
Three lone pairs are at equatorial position to minimise repulsion.
Thus, (iv) → (p)

(i) OSF₂ 

sp³-hybridised S -atom (Ip - bp) repulsion, trigonal pyramidal
(ii)→ (s)
(ii) O₂SF₂

 
sp³-hybridised s-atom tetrahedral
(ii) → (P)
(iii) XeF₄

sp³d²-hybridised Xe-atom.
Two lone pairs, square planar (iii) → (r)

sp³-hybridised Cl atom, tetrahedral.
(iv) → (p)

sp³d-hybridised I, three lone pairs at equatorial position linear,
(v) → (q)

(i) (0)

(Ip-bp) repulsion will cause distortion and thus, bond angle < 90° in each.

Thus, three lone pairs at equatorial position to minimise repulsion.

(iii) (6) Melamine is heterocyclic basic com pound (six N-atoms) each N having one lone pair.