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Which is paramagentic?

$C l_{2} O_{6}$

$C l_{2} O_{7}$

$C l_{2} O$

$C l O_{2}$

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Explanation

(d) $C l O_{2}$ has 33 electrons ; one will be unpaired

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Greater the dipole moment:

greater is the ionic nature

lesser the polarity

smaller is the ionic nature

none of these

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Explanation

(a) more is the dipole moment more is ionic nature. $μ = δ \times d;$ higher is $μ$ , more will be $δ$ on that atom.

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$N F_{3}$ is:

non-polar compound

electrovalent compound

having low value of dipole moment than $N H_{3}$

having more dipole moment than $N H_{3}$

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Explanation

(c) Inspite of three polar bond,the lone pair of electrons on N atom decreases the dipole moment of $N F_{3}$ than $N H_{3}$ because in NH3 , dipole moment is getting added ,where as in NF3, dipole moment is being substracted.

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Polarization of electrons in acrolein may be written as:

$\overset{δ^{-}}{C} H_{2} = C H - \overset{δ^{+}}{C} H = O$

$\overset{δ^{-}}{C} H_{2} = C H - C H = \overset{δ^{+}}{O}$

$\overset{δ^{-}}{C} H_{2} = \overset{δ^{+}}{C} H - C H = O$

$\overset{δ^{+}}{C} H_{2} = C H - C H = \overset{δ^{-}}{O}$

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Explanation

(d) O is more electronegative than C.

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$P C l_{5}$ exists but $N C l_{5}$ does not because:

nitrogen has no vacant 2d-orbitals

$N C l_{5}$ is unstable

nitrogen atom is much smaller than p

nitrogen is highly inert

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Explanation

(a) Thus, excitation of 2s-electron in N is not possible.and N doesnot have vacent d sub shell.

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Bond angle between two hybrid orbitals is $105^{°}$ .% S character orbital is:

between 30-31%

between 19-20%

between 25 -26%

between 22-23%

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Explanation

(d) Nearer to 25%, the angle is $109^{°}$ 28' for $s p^{3}$ -hybridization.

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Which bond angle would result in the maximum dipole moment for the triatomic molecule $X Y_{2}$ shown below?

$90^{°}$

$120^{°}$

$150^{°}$

$180^{°}$

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Explanation

(a) $μ = \sqrt{{μ_{1}}^{2} + {μ_{2}}^{2} + μ_{1} μ_{2} \cos θ}$ , if $θ = 90^{°}$ $μ$ is maximum.

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The hydrogen bonding is strongest in:

$O — H ‐ ‐ ‐ S$

$S — H ‐ ‐ ‐ O$

$F — H ‐ ‐ ‐ F$

$F — H ‐ ‐ ‐ O$

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Explanation

$‐ ‐ ‐ H — F > ‐ ‐ ‐ H — O > ‐ ‐ ‐ H — N$

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The ionisation energy will be maximum for the process:

$B a \to B a^{2 +}$

$B e \to B e^{2 +}$

$C s \to C s^{+}$

$L i \to L i^{+}$

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Explanation

(b) smaller is atom, more is energy needed to remove electron , i.e. ionisation energy. Also removal of two electrons needs more energy.

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In which of the process, the bond order increases and magnetic behaviour changes?

$N_{2} \to {N_{2}}^{+}$

$C_{2} \to {C_{2}}^{+}$

$N O \to N O^{+}$

$O_{2} \to {O_{2}}^{+}$

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Explanation

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