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CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 14 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
The P-Block Elements
Quiz 14
The electronic configuration of chlorine is _________
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2, 7
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2, 8, 8, 7
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2, 8, 7
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2, 7, 8
In which of the following, oxidation number of chlorine is $$ + 5 ?$$
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$$Cl_{2}O_{7}$$
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$$\mathrm { ClO } _ { 3 } ^ { - }$$
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$$\mathrm { CHO } ^ { - }$$
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$$\mathrm { ClO } _ { 4 } ^ { - }$$
Explanation
Oxidation number of $$Cl_2O_7$$:-
$$2x-14=0$$
$$x=7$$
Oxidation number of $${ClO_3}^{-}$$:-
$$x-6=-1$$
$$x=5$$
Oxidation number of $$ClO^-$$:-
$$x-2=-1$$
$$x=1$$
Oxidation number of $${ClO_4}^-$$:-
$$x-8=-1$$
$$x=7$$
hence option B is correct.
Which of the following compounds undergoes haloform reaction?
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$$CH_{3}COCH_{3}$$
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$$HCHO$$
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$$CH_{3}CH_{2}Br$$
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$$CH_{3} - O - CH_{3}$$
Explanation
The haloform reaction comprises the reaction of a halogen with a methyl ketone (a molecule containing the $$R–CO–CH_3$$ grouping) in the presence of a base.
Among the given compound, propanone undergoes haloform reaction as it contains a methyl group attached to carbonyl carbon.
Hence, option A is correct.
Which of the following is correct statement regarding $$N(CH_3)_3$$ and $$N(SiH_3)_3$$?
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$$N(SiH_3)_3$$ is planar and more basic than $$N(CH_3)_3$$
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$$N(SiH_3)_3$$ is planar and less basic than $$N(CH_3)_3$$
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$$N(CH_3)_3$$ is planar and more basic than $$N(SiH_3)_3$$
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$$N(CH_3)_3$$ is pyramidal and less basic than $$N(SiH_3)_3$$
Explanation
$$N(SiH_3)_3$$ is planar due to back bonding and $$N(CH_3)_3$$ more basic due to presence of lone pair electrons.
What is the action of following reagents on ammonia ?
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i)Excess of air
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ii)Excess of chlorine
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iii)Na metal
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none of these
Which of the following is incorrect?
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Red $$P$$ is toxic
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White $$'P'$$ is highly soluble in $$CS_{2}$$
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Black $$'P'$$ is thermodynamic is most stable
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White $$'P'$$ is soluble in $$NaOH$$ evolves $$PH_{3}$$
Explanation
Red P is not considered as toxic in its pure form.
Which of the following is true for $$N_{2}O_{5}$$?
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Paramagnetic
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Anhydride of $$HNO_{2}$$
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Brown gas
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Exist in solid In form of $$[NO^{+}_{2}][NO_{3}^{-}]$$
Explanation
$$N_{2}O_{5}$$ in solid form exists as $$NO_{3}^{-}$$ and $$NO_{2}^{-}$$.
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