Chemical Bonding and Molecular Structure Test 51

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Chemical Bonding and Molecular Structure Test 51

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Weekly Quiz Competition

Question 1
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Which of the following represent $$sp - sp - sp^2 - sp - sp^2 - sp^3$$ hybridization from left to right order?

A
$$CH_2= CH - C = \overset{H}{\overset{|}{C}} - C \equiv N$$

B
$$CH \equiv C - CH = C = CH-CH_3$$

C
$$N \equiv C - \overset{H}{\overset{|}{C}} - \underset{H}{\underset{|}{C}} - CH - CH_3$$

D

Solution

Carbon having single bond is $$sp^3$$ hybridised, carbon with double bond is $$sp^2$$ hybridised and the carbon having triple bond is $$sp$$ hybridised.
In option B $$CH$$ is $$sp$$ hybridised, $$C$$ is $$sp$$ hybridised, $$CH$$ is $$sp^2$$ hybridised, $$C$$ is $$sp$$ hybridised, $$CH$$ is $$sp2$$ hybridised and $$CH_3$$ is $$sp^3$$ hybridised.
Question 2
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A molecule $$(X)$$ has (i) four sigma bonds formed by the overlap of $$sp^{2}$$ and $$s$$ orbitals; (ii) one sigma bond formed by $$sp^{2}$$ and $$sp^{2}$$ - orbitals and (iii) one $$\pi-bond$$ formed by $$p_{z}$$ and $$p_{z}$$ orbitals. Which of the following is $$X$$?

A
$$C_{2}H_{6}$$

B
$$C_{2}H_{5}Cl$$

C
$$C_{2}H_{2}Cl_{2}$$

D
$$C_{2}H_{4}$$

Solution
Question 3
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Select the correct statement :

A
If molecule has any polar bond, then it is always polar

B
Solubility of noble gases increases in water down the group when their size increases because London dispersion force increases

C
First ionization energy of Al is greater than that of gallium

D
$$ XeF^{ \oplus}_5 $$ has disorted octahedral geometry

Solution
Question 4
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The bond order of $$ O_2^+ $$ is:

A
$$1$$

B
$$1.5$$

C
$$2.5$$

D
$$3$$

Solution

Hint: Bond order can be calculated by using molecular orbital theory.
Step 1: Determination of Bond Order :
Total number of electrons in $$O_2=16$$ but due to a positive charge one electron is deducted and thus,
the total number of electrons in $${O_2}^+ = 15$$
Molecular orbital configuration can be written as,
$$(\sigma 1s)^2 <(\sigma\ast 1s)^2< (\sigma 2s)^2< (\sigma\ast 2s)^2< (\sigma 2pz)^2< (\pi 2px)^2 \approx (\pi 2py)^2< (\pi\ast 2px)^1 \approx (\pi\ast 2py)^0 $$
$$\ast \ -$$ mark represent the electron is present in anti-bonding molecular orbital.
Bond order can be defined as:
$$Bond\ Order = \dfrac{Number\ of\ electrons\ in\ B.M.O\ –\ Number\ of\ electrons\ in\ A.B.M.O.}{2} $$
( where, B.M.O. - Bonding Molecular Orbital, A.B.M.O - Anti-bonding Molecular Orbital )
$${O_2}^+$$ has $$10$$ bonding electrons and $$5$$ antibonding electrons.
So, the bond order is $$\dfrac{10-5}{2}=2.5$$.
Final Step: Correct Answer - (C) $$2.5$$.
Question 5
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Find out the similarities between $$ I_2Cl_6 $$ and $$ Al_2Cl_6 .$$

A
Both have $$3\ C -4e^-$$ bond

B
Both have $$ sp^3 $$ hybridization for the central atom

C
Both are non planar

D
All are correct

Solution

a) $$I_2Cl_6$$
Hybridization of each $$ I : sp^3 d^2 $$
Planar structure
2 bonds are of $$3C -4e^-$$
4 bonds are of $$ 2C - 2 e^- $$

b) $$Al_2Cl_6$$
Hybridization of each $$ Al : sp^3 $$
Non-polar structure
2 bonds are of $$ 3C - 4e^- $$
4 bonds are of $$ 2C - 2e^- $$
Question 6
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Which of the following molecules/ions are all the bonds not equal?

A
$$ SF_4 $$

B
$$ SiF_4 $$

C
$$ XeF_4 $$

D
$$ BF_4^- $$

Solution

In $$ SF_4 ,$$ the central aton S has $$ sp^d $$-hybridization. Thus it contains two axial and two equatorial bonds to give a see-saw structure .
Question 7
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Back bonding always changes

A
bond angle

B
hybridisation of central atom

C
planarity

D
bond length

Solution
Question 8
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Directions For Questions

Hybridization is the measure to express the molecular shapes reasonably. Hybridization, although theoretical, but is highly correlated with molecular shapes or inter orbital angles. If there is any departure in geometry in apparent departure in hybridization can be observed and the following characteristic relationship is the easiest way to interpret that departure. Any hybrid orbital made by s and p-orbitals is characterized by the following relationship.

$$ f_s = \dfrac {1}{i + 1}$$ or $$f_p = \dfrac {i}{i+1}$$ and $$i =\dfrac {f_p}{f_s} $$

For $$ sp^3, i= 3$$ and $$sp^2, i=2 $$ for an orbital, $$ f_s + f_p = 1 $$ while all the s-p hybrids of gein atom must satisfy the condition.

$$ \sum { f_x} = 1.00 $$

d and the inter orbital angle between two equivalent ( same $$f_s$$ and same $$f_p$$ ) hybrid orbitals is given by

$$ cos \theta = \dfrac {-1}{i} $$

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VSEPR theory suggest a small departure from $$ 109^0 28'$$ for $$ NH_3 $$ molecule. The H-N-H bond angle has been found to be $$ 107^0 $$ what kind of hybrid orbital will be occupied by the non -bonding pair of electron?

A
$$ sp^3 $$

B
$$ sp^{3.4} $$

C
$$ sp^{2.13} $$

D
$$ sp^2 $$

Solution
Question 9
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Which of the following has -O-O- linkage ?

A
$$ H_2S_2O_6 $$

B
$$ H_2S_2O_8 $$

C
$$ H_2S_2O_3 $$

D
$$ H_2S_4O_6 $$

Solution
Question 10
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Which species has the maximum number of lone pair of electrons on the central atom ?

A
$$ [ ClO^-_3] $$

B
$$ XeF_4 $$

C
$$ SF_4 $$

D
$$ [ I^-_3] $$

Solution