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1
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { From the following pair of compounds, identify the incorrect pair in terms of the covalent character. } $$

A
$$ \mathrm{CuS}>\mathrm{CuO} $$
B
$$ \mathrm{BeCl}_2>\mathrm{MgCl}_2 $$
C
$$ \mathrm{SnCl}_4>\mathrm{SnCl}_2 $$
D
$$ A g C l>A g I $$
2
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

The bond dissociation enthalpy of the species in their correct order is:

A
$$\mathrm{O}_2^{2-}>\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}$$
B
$$\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}$$
C
$$\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}$$
D
$$\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}$$
3
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

In the following question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below.

Assertion (A): A molecule of $$\mathrm{SF}_4$$ is see-saw shaped, while that of $$\mathrm{CIF}_3$$ is T-shaped.

Reason(R): SF$$_4$$ has two lone pair of electrons. But CIF$$_3$$ has one pair of electrons.

A
Assertion (A) is incorrect but Reason (R) is correct
B
Assertion (A) and Reason (R) are correct
C
Both Assertion (A) and Reason (R) are incorrect
D
Assertion (A) is correct but Reason (R) is incorrect
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

On the basis of VSEPR theory, match the molecules listed in Column I with their shapes given in Column II.

No. Column I No. Column II
A $$\mathrm{ClF_3}$$ P Sea saw
B $$\mathrm{BrF_5}$$ Q Pentagonal bipyramidal
C $$\mathrm{SF_4}$$ R Square pyramidal
D $$\mathrm{IF_7}$$ S T-shaped

A
$$ \mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q} $$
B
$$ \mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q} $$
C
$$ \mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q} $$
D
$$ \mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{R} $$
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