CUET Chemistry Set-3: Inorganic Chemistry - Blocks
Question 1
Which of the following oxides is most acidic?
(1) $\mathrm{N_2O_5}$ (2) $\mathrm{P_4O_{10}}$ (3) $\mathrm{As_4O_6}$ (4) $\mathrm{Sb_4O_6}$
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Answer: (1)
Solution: Acidic character decreases down the group; nitrogen is the lightest p-block element listed, so $\mathrm{N_2O_5}$ is the most acidic oxide.
Question 2
The stability of the +1 oxidation state in group 13 follows the order
(1) $\mathrm{Al} > \mathrm{Ga} > \mathrm{In} > \mathrm{Tl}$ (2) $\mathrm{Tl} > \mathrm{In} > \mathrm{Ga} > \mathrm{Al}$ (3) $\mathrm{Ga} > \mathrm{Al} > \mathrm{In} > \mathrm{Tl}$ (4) $\mathrm{Al} > \mathrm{Tl} > \mathrm{In} > \mathrm{Ga}$
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Answer: (2)
Solution: Inert-pair effect increases down the group; hence +1 state becomes more stable from Al to Tl.
Question 3
Borax bead test is given by
(1) $\mathrm{Al^{3+}}$ (2) $\mathrm{Mg^{2+}}$ (3) $\mathrm{Cu^{2+}}$ (4) $\mathrm{Na^+}$
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Answer: (3)
Solution: Transition-metal ions (here $\mathrm{Cu^{2+}}$) give coloured borax beads; s-block ions do not.
Question 4
The compound that does NOT exist is
(1) $\mathrm{BF_3}$ (2) $\mathrm{BCl_3}$ (3) $\mathrm{BI_3}$ (4) $\mathrm{BH_5}$
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Answer: (4)
Solution: Boron can expand its octet to some extent but $\mathrm{BH_5}$ violates stable hypervalency; it is unknown.
Question 5
Graphite conducts electricity due to
(1) $\sigma$ electrons (2) $\pi$ electrons (3) ionic mobility (4) dangling bonds
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Answer: (2)
Solution: Delocalised $\pi$ electrons within layered sheets act as charge carriers.
Question 6
The correct order of catenation is
(1) $\mathrm{C} > \mathrm{Si} > \mathrm{Ge} \approx \mathrm{Sn}$ (2) $\mathrm{Si} > \mathrm{C} > \mathrm{Ge} > \mathrm{Sn}$ (3) $\mathrm{Ge} > \mathrm{Sn} > \mathrm{Si} > \mathrm{C}$ (4) $\mathrm{Sn} > \mathrm{Ge} > \mathrm{Si} > \mathrm{C}$
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Answer: (1)
Solution: Catenation decreases with increasing atomic size; carbon forms the longest chains.
Question 7
Which allotrope of phosphorus is most reactive?
(1) Red (2) Black (3) White (4) Violet
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Answer: (3)
Solution: White phosphorus consists of discrete $\mathrm{P_4}$ molecules with strained 60° angles, making it highly reactive.
Question 8
The basicity of $\mathrm{NH_3}$ is higher than that of $\mathrm{PH_3}$ because
(1) electronegativity of N > P (2) lone-pair on N is more localized (3) hydrogen bonding in $\mathrm{NH_3}$ (4) larger size of P
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Answer: (1)
Solution: Higher electronegativity of N increases electron density on the lone pair, enhancing proton acceptance.
Question 9
The gas evolved on heating ammonium dichromate is
(1) $\mathrm{NH_3}$ (2) $\mathrm{N_2}$ (3) $\mathrm{N_2O}$ (4) $\mathrm{NO_2}$
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Answer: (2)
Solution: $(\mathrm{NH_4})_2\mathrm{Cr_2O_7} \rightarrow \mathrm{N_2} + 4\mathrm{H_2O} + \mathrm{Cr_2O_3}$; green $\mathrm{Cr_2O_3}$ and $\mathrm{N_2}$ gas are formed.
Question 10
Which of the following does NOT show disproportionation?
(1) $\mathrm{H_3PO_3}$ (2) $\mathrm{HNO_2}$ (3) $\mathrm{H_2SO_4}$ (4) $\mathrm{Cl_2O_7}$
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Answer: (3)
Solution: $\mathrm{H_2SO_4}$ contains sulphur in its highest oxidation state (+6), hence cannot disproportionate.
Question 11
The correct statement for interhalogen compounds is
(1) $\mathrm{BrF_5}$ is square pyramidal (2) $\mathrm{ClF_3}$ is T-shaped (3) $\mathrm{IF_7}$ is pentagonal bipyramidal (4) All of these
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Answer: (4)
Solution: All three geometries are correctly assigned based on VSEPR theory.
Question 12
The only halogen that forms a stable hypohalous acid in aqueous solution is
(1) F (2) Cl (3) Br (4) I
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Answer: (1)
Solution: OF$_2$ is not hypohalous, but HOF (hypofluorous acid) is known only for fluorine; however, among true hypohalous acids (HOX), HOCl is most stable; yet strictly speaking HOF is unique to F. Recast: the question intends the halogen whose HOX is best known—(2) Cl. (Corrected interpretation)
Revised Answer: (2)
Solution: HOCl is the most stable hypohalous acid; fluorine forms HOF only transiently.
Question 13
XeF$_2$ reacts with $\mathrm{H_2O}$ to give
(1) $\mathrm{Xe} + \mathrm{HF} + \mathrm{O_2}$ (2) $\mathrm{XeO_3} + \mathrm{HF}$ (3) $\mathrm{XeOF_2} + \mathrm{H_2}$ (4) $\mathrm{Xe} + \mathrm{HF} + \mathrm{O_3}$
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Answer: (1)
Solution: $2\mathrm{XeF_2} + 2\mathrm{H_2O} \rightarrow 2\mathrm{Xe} + 4\mathrm{HF} + \mathrm{O_2}$; xenon is liberated with oxygen.
Question 14
The magnetic moment of $\mathrm{K_2[NiF_6]}$ is closest to
(1) 0 BM (2) 1.73 BM (3) 2.83 BM (4) 3.87 BM
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Answer: (3)
Solution: Ni(IV) in $\mathrm{K_2[NiF_6]}$ is $d^6$ low-spin octahedral with two unpaired electrons; $\mu = \sqrt{2(2+2)} = 2.83$ BM.
Question 15
Which ion is colourless?
(1) $\mathrm{Ti^{3+}}$ (2) $\mathrm{Sc^{3+}}$ (3) $\mathrm{V^{3+}}$ (4) $\mathrm{Cr^{3+}}$
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Answer: (2)
Solution: $\mathrm{Sc^{3+}}$ is $d^0$; no d–d transitions → colourless.
Question 16
The lanthanide contraction is responsible for
(1) similar radii of Zr and Hf (2) high density of Os (3) basicity of La(OH)$_3$ (4) paramagnetism of Gd
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Answer: (1)
Solution: Lanthanide contraction makes 5d and 4d congeners (Zr/Hf) almost identical in size.
Question 17
The strongest reducing agent among the following is
(1) $\mathrm{Eu^{2+}}$ (2) $\mathrm{Yb^{2+}}$ (3) $\mathrm{Sm^{2+}}$ (4) $\mathrm{Ce^{2+}}$
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Answer: (1)
Solution: $\mathrm{Eu^{2+}}$ has the most negative $E^\circ$ value; stable $f^7$ configuration after oxidation makes it the best reducer.
Question 18
The ore that is NOT concentrated by froth flotation is
(1) $\mathrm{CuFeS_2}$ (2) $\mathrm{PbS}$ (3) $\mathrm{Al_2O_3·2H_2O}$ (4) $\mathrm{ZnS}$
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Answer: (3)
Solution: Bauxite is an oxide ore; froth flotation suits sulphides.
Question 19
In the extraction of Al, the function of cryolite is
(1) to lower the melting point of $\mathrm{Al_2O_3}$ (2) to act as a flux (3) to increase conductivity (4) all of these
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Answer: (4)
Solution: Cryolite ($\mathrm{Na_3AlF_6}$) performs all three roles in Hall–Héroult cell.
Question 20
The metal that forms a volatile carbonyl is
(1) Fe (2) Ni (4) Co (4) all of these
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Answer: (4)
Solution: Fe, Ni, Co all form stable volatile carbonyls used in Mond process.
Question 21
The catalyst used in Haber’s process is
(1) $\mathrm{Fe} + \mathrm{K_2O} + \mathrm{Al_2O_3}$ (2) $\mathrm{V_2O_5}$ (3) $\mathrm{Pt}$ (4) $\mathrm{Ni}$
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Answer: (1)
Solution: Finely divided Fe promoted with $\mathrm{K_2O}$ and $\mathrm{Al_2O_3}$ is the industrial catalyst.
Question 22
The geometry of $\mathrm{[Ni(CO)_4]}$ is
(1) square planar (2) tetrahedral (3) octahedral (4) trigonal bipyramidal
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Answer: (2)
Solution: Ni(0) $d^{10}$ with 4 CO ligands gives $sp^3$ tetrahedral geometry.
Question 23
The complex ion that is diamagnetic is
(1) $\mathrm{[Fe(CN)_6]^{3-}}$ (2) $\mathrm{[CoF_6]^{3-}}$ (3) $\mathrm{[Ni(CN)_4]^{2-}}$ (4) $\mathrm{[MnCl_4]^{2-}}$
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Answer: (3)
Solution: $\mathrm{[Ni(CN)_4]^{2-}}$ is square planar $dsp^2$ with Ni(II) $d^8$ low-spin → no unpaired electrons.
Question 24
The number of unpaired electrons in $\mathrm{[Cr(H_2O)_6]^{3+}}$ is
(1) 1 (2) 2 (3) 3 (4) 4
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Answer: (3)
Solution: Cr(III) $d^3$ in weak octahedral field; $t_{2g}^3 e_g^0$ → 3 unpaired electrons.
Question 25
The ligand that can show linkage isomerism is
(1) $\mathrm{NH_3}$ (2) $\mathrm{H_2O}$ (3) $\mathrm{NO_2^-}$ (4) $\mathrm{Cl^-}$
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Answer: (3)
Solution: $\mathrm{NO_2^-}$ can bind through N (nitro) or O (nitrito), hence exhibits linkage isomerism.
BETA
