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CHEMISTRY(KCET)

285266 The hybridisation of C in diamond, graphite andethyne is in the order

1

s p^{3}, s p, s p^{3}

2

s p^{3}, s p^{2}, s p

3

s p, s p^{2}, s p^{3}

4

s p^{2}, s p^{3}, s p

Karnataka CET 2016

CHEMISTRY(KCET)

285267 Using MOT, compare $O_{2}^{+}$ and $O_{2}^{-}$ species and choose the incorrect option

1

O_{2}^{-}

is less stable.

2 Both

O_{2}^{+}

and

O_{2}^{-}

are paramagnetic.

3

O_{2}^{+}

have higher bond order than

O_{2}^{-}

.

4

O_{2}^{+}

is diamagnetic while

O_{2}^{-}

is paramagnetic.

Explanation:

(d) Molecular orbital configuration of $O_{2}^{+}$ :
$σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ * 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2}$
$= π 2 p_{y}^{2} π^{*} 2 p_{x}^{1}$
Bond order $= \frac{1}{2} (10 - 5) = 2.5$
Number of unpaired electrons $= 1$ , hence paramagnetic.
Molecular orbital configuration of $O_{2}^{-}$ :
$σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2}$
$= π 2 p_{y}^{2} π^{*} 2 p_{x}^{2} = π^{*} 2 p_{y}^{1}$
Bond order $= \frac{1}{2} (10 - 7) = 1.5$
Number of unpaired electron $= 1$ , hence paramagnetic.
Both $O_{2}^{+}$ and $O_{2}^{-}$ are paramagnetic.
$O_{2}^{+}$ have higher bond order than $O_{2}^{-}$ .
$O_{2}^{-}$ is less stable.

Karnataka CET 2015

CHEMISTRY(KCET)

285268 The pair of compounds having identical shapes for their molecules is

1

{CH}_{4}, {SF}_{4}

2

{BCl}_{2}, {ClF}_{3}

3

{XeF}_{2}, {ZnCl}_{2}

4

{SO}_{2}, {CO}_{2}

Explanation:

(c) ${XeF}_{2}$ is linear in shape with 2 bond pairs and 3 lone pairs of electrons.
${ZnCl}_{2}$ is also linear in shape with 2 bond pairs of electrons.

Karnataka CET 2014

CHEMISTRY(KCET)

285266 The hybridisation of C in diamond, graphite andethyne is in the order

1

s p^{3}, s p, s p^{3}

2

s p^{3}, s p^{2}, s p

3

s p, s p^{2}, s p^{3}

4

s p^{2}, s p^{3}, s p

Karnataka CET 2016

CHEMISTRY(KCET)

285267 Using MOT, compare $O_{2}^{+}$ and $O_{2}^{-}$ species and choose the incorrect option

1

O_{2}^{-}

is less stable.

2 Both

O_{2}^{+}

and

O_{2}^{-}

are paramagnetic.

3

O_{2}^{+}

have higher bond order than

O_{2}^{-}

.

4

O_{2}^{+}

is diamagnetic while

O_{2}^{-}

is paramagnetic.

Explanation:

(d) Molecular orbital configuration of $O_{2}^{+}$ :
$σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ * 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2}$
$= π 2 p_{y}^{2} π^{*} 2 p_{x}^{1}$
Bond order $= \frac{1}{2} (10 - 5) = 2.5$
Number of unpaired electrons $= 1$ , hence paramagnetic.
Molecular orbital configuration of $O_{2}^{-}$ :
$σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2}$
$= π 2 p_{y}^{2} π^{*} 2 p_{x}^{2} = π^{*} 2 p_{y}^{1}$
Bond order $= \frac{1}{2} (10 - 7) = 1.5$
Number of unpaired electron $= 1$ , hence paramagnetic.
Both $O_{2}^{+}$ and $O_{2}^{-}$ are paramagnetic.
$O_{2}^{+}$ have higher bond order than $O_{2}^{-}$ .
$O_{2}^{-}$ is less stable.

Karnataka CET 2015

CHEMISTRY(KCET)

285268 The pair of compounds having identical shapes for their molecules is

1

{CH}_{4}, {SF}_{4}

2

{BCl}_{2}, {ClF}_{3}

3

{XeF}_{2}, {ZnCl}_{2}

4

{SO}_{2}, {CO}_{2}

Explanation:

(c) ${XeF}_{2}$ is linear in shape with 2 bond pairs and 3 lone pairs of electrons.
${ZnCl}_{2}$ is also linear in shape with 2 bond pairs of electrons.

Karnataka CET 2014

CHEMISTRY(KCET)

285266 The hybridisation of C in diamond, graphite andethyne is in the order

1

s p^{3}, s p, s p^{3}

2

s p^{3}, s p^{2}, s p

3

s p, s p^{2}, s p^{3}

4

s p^{2}, s p^{3}, s p

Karnataka CET 2016

CHEMISTRY(KCET)

285267 Using MOT, compare $O_{2}^{+}$ and $O_{2}^{-}$ species and choose the incorrect option

1

O_{2}^{-}

is less stable.

2 Both

O_{2}^{+}

and

O_{2}^{-}

are paramagnetic.

3

O_{2}^{+}

have higher bond order than

O_{2}^{-}

.

4

O_{2}^{+}

is diamagnetic while

O_{2}^{-}

is paramagnetic.

Explanation:

(d) Molecular orbital configuration of $O_{2}^{+}$ :
$σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ * 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2}$
$= π 2 p_{y}^{2} π^{*} 2 p_{x}^{1}$
Bond order $= \frac{1}{2} (10 - 5) = 2.5$
Number of unpaired electrons $= 1$ , hence paramagnetic.
Molecular orbital configuration of $O_{2}^{-}$ :
$σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2}$
$= π 2 p_{y}^{2} π^{*} 2 p_{x}^{2} = π^{*} 2 p_{y}^{1}$
Bond order $= \frac{1}{2} (10 - 7) = 1.5$
Number of unpaired electron $= 1$ , hence paramagnetic.
Both $O_{2}^{+}$ and $O_{2}^{-}$ are paramagnetic.
$O_{2}^{+}$ have higher bond order than $O_{2}^{-}$ .
$O_{2}^{-}$ is less stable.

Karnataka CET 2015

CHEMISTRY(KCET)

285268 The pair of compounds having identical shapes for their molecules is

1

{CH}_{4}, {SF}_{4}

2

{BCl}_{2}, {ClF}_{3}

3

{XeF}_{2}, {ZnCl}_{2}

4

{SO}_{2}, {CO}_{2}

Explanation:

(c) ${XeF}_{2}$ is linear in shape with 2 bond pairs and 3 lone pairs of electrons.
${ZnCl}_{2}$ is also linear in shape with 2 bond pairs of electrons.

Karnataka CET 2014

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