Chemical Bonding and Molecular Structure Test 7

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Chemical Bonding and Molecular Structure Test 7

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1-16) This section contains 16 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q.
The shape [CIF₄]^- and [ClF₂]^- ions is respectively

A
see-saw and linear

B
see-saw and bent

C
tetrahedral and linear

D
square planar and linear

Solution
Question 2
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Which of the following sequences shows the correct bond angle order for isoelectronic species

A

B

C

D
Cannot be predicted

Solution

In each central atom is sp²-hybridised, each having one lone pair.
Difference in the extent of repulsion changes value of θ.
Question 3
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XeF₂ is isostructural with

A
TeF₂

B

C
SbCI₃

D
BaCl₂

Solution

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 .
Question 4
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The species having pyramidal shape is
[IIT JEE 2010]

A
SO₂

B
BrF₃

C

D
OSF₂

Solution

(b) BrF₃ - T-shaped (sp³d²)
Question 5
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Molecular shape of CF₄, SF₄ and XeF₄ are

A
the same, with 2, 0 and 1 lone pair of electrons respectively

B
the same with 1, 1 and 1 lone pair of electrons

C
different, with 0, 1 and 2 lone pair of electrons

D
different, with 1, 0 and 2 lone pair of electrons

Solution
Question 6
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Two of the following species have same shape

A

B

C

D

Solution

Both have one lone pair-trigonal pyramidal.
Question 7
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ECI₃ (where, E = B, P, As, Bi) of these elements are known.
Bond angles are in the following order

A
B > P> As > Bi

B
B > P = As > Bi

C
Bi > As > P > B

D
B > As > Bi > P

Solution

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.
Question 8
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Consider the two structures :
Select the correct statement(s).

A
I is bent, II is linear

B
I is linear, II is bent

C
Both is bent

D
Both are linear

Solution

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.
Question 9
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Which of the two ions from the list given below that have the geometry that is explained by the same hybridisation of orbitals?

A

B

C

D

Solution
Question 10
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The correct order of increasing bond angles in the following species is

A
CIO^-₂ -₃ CIO^-₄

B
CIO^-₂ < CIO^-₄ CIO^-₃

C
CIO^-₄ < CIO^-₃ CIO^-₂

D
equal since each Cl atom is sp³-hybridised

Solution
Question 11
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Which of the following molecule/species has the minimum number of lone pairs?

A
ICI₃

B
BF₄^-

C
SnCI₂

D
XeF₄

Solution
Question 12
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PCI₅ has a shape of trigonal bipyramid whereas, IF₅ has a shape of square pyramid. It is due to

A
presence of unshared electron pair on I which is oriented so as to minimise repulsion while in PCI₅, P has no non-bonding pair

B
octet of P is complete that of I is incomplet

C
P and I are of different groups and periods

D
extent of different repulsion in F and I

Solution

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.
Question 13
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For H₂O and H₂S, given
Q.
Difference in bond angle is due to

A
the larger size of S-atom as compared to O-atom which minimises repulsion and allows the bonds in H₂S to be purely p-type

B
liquid state of H₂O as compared to gaseous state of H₂S

C
Both of the above

D
None of the above

Solution

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.
Question 14
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For which of the following sets of geometry, both axial and equatorial positions are present?

A
Octahedral, trigonal bipyramidal

B
Tetrahedral, pentagonal bipyramidal

C
Trigonal bipyramidal, pentagonal bipyramidal

D
Tetrahedral, octahedral

Solution
Question 15
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Which of the following set of molecules have the same shape but different hybridisation?

A
H₂O, H₂S

B
XeO₃, BrF₃

C
Ni(CN)4]^2-, [Ni(CO)₄]

D
BeCI2, l-₃

Solution
Question 16
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Which is the correct order of the bond angle?

A
NH₃ < NF₃

B
H₂O > Cl₂O

C
PH₃< SbH₃

D
H₂Te < H₂S

Solution

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.
Question 17
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Direction (Q. Nos. 21-22) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).

Following reaction, CIF₃ + AsF5 → (CIF₂⁺ ) (AsF^-₆)
Q.
Select the correct statements.

A
Hybridisation of Cl changes from sp² to sp

B
Hybridisation of As changes from sp³d to sp³d²

C
Both (a) and (b) are correct

D
None of the above is correct

Solution

sp³d -hybridised Cl-atom - (T-shaped s tructure)
sp³ -hybridised. Cl-atom (Bent)
sp³d -hybridised As-atom (Trigonal bipyramidai).
sp³d² -hybridised As-atom (Octahedral).
Question 18
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Following reaction, CIF₃ + AsF5 → (CIF₂⁺ ) (AsF₆^-)
Q.
Select the correct statement(s).

A
CIF₃ changes its structure from pyramidal to angular in CIF⁺₂

B
AsF₅ changes to structure from angular bipyramidal to octahedral in (AsF₆)^-

C
Both (a) and (b) are correct

D
None of the above is correct

Solution

sp³d -hybridised Cl-atom - (T-shaped s tructure)
sp³ -hybridised. Cl-atom (Bent)
sp³d -hybridised As-atom (Trigonal bipyramidai).
sp³d² -hybridised As-atom (Octahedral).
Question 19
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Direction (Q. Nos. 23 and 24) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q.
Match the species in Column I with the structure in Column II.

A
a

B
b

C
c

D
d

Solution

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)
Question 20
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Match the species in Column I with the shape in Column II.

A
a

B
b

C
c

D
d

Solution

(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)