Match the catalysts (Column I) with products column (II) Column I Column II (A) $${V}_{2}{O}_{5}$$ (i) Polyethylene (B) $$Ti{Cl}_{4}/Al{(Me)}_{3}$$ (ii) ethanal (C) $$Pd{Cl}_{2}$$ (iii) $${H}_{2}{SO}_{4}$$ (D) Iron Oxide (iv) $${NH}_{3}$$

(A)- (iii); (B)- (i); (C)- (ii); (D)- (iv)

(A)- (ii); (B)- (iii); (C)- (i); (D)- (iv)

(A)- (iii); (B)- (iv); (C)- (i); (D)- (ii)

(A)- (iv); (B)- (iii); (C)- (ii); (D)- (i)

Match the following catalyst with their products- Sr Column 1 Sr Column 2 1. $$TiCl_4+AlCl_3$$ a. Ethanal 2. $$V_2O_5$$ b. $$NH_3$$ 3. $$Fe$$ c. Polyethylene 4. $$Pd$$ d. $$H_2SO_4$$

$$(i)-C ; (ii)-D ; (iii)-B ; (iv)-A$$

$$(i)-B ; (ii)-A ; (iii)-D ; (iv)-C$$

$$(i)-D ; (ii)-A ; (iii)-B; (iv)-C$$

$$(i)-A ; (ii)-B ; (iii)-C ; (iv)-D$$

MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 13

Identify the incorrect statement in the following

0%
d block elements show irregular and erratic chemical properties among themselves.
0%
La and Lu have partially filled d-orbital and no other partially filled orbital.
0%
The chemistry of various Lanthanoids is similar.
0%
4f and 5f orbitals are equally shielded.

The thermal decomposition of ammonium dichromate produces a green residue of:

0%
$Cr_2O_3$
0%
$CrO_2$
0%
$CrO_3$
0%
$CrO_5$

Number of identical

$Cr-O$ bonds in dichromate ion

${Cr_2{ O }_{ 7 }}^{ 2- }$ is:

0%
4
0%
6
0%
7
0%
8

$\left[ Cr\left( { H }_{ 2 }O \right) _{ 6 }{ Cl }_{ 3 } \right]$ has a magnetic moment of 3.83 BM. The correct distribution of 3d electrons in the chromium atom in the complex is:

0%
${ 3d }_{ xy }^{ 1 },{ 3d }_{ yz }^{ 1 },{ 3d }_{ { z }^{ 2 } }^{ 1 }$
0%
${ 3d }_{ \left( { x }^{ 2 }-{ y }^{ 2 } \right)} ^1 ,{ 3d }_{ z^2 }^{ 1 },{ 3d }_{ xz }^{ 1 }$
0%
${ 3d }_{ xy }^{ 1 },{ 3d }_{ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^1,{ 3d }_{ z }^{ 1 }$
0%
${ 3d }_{ xy }^{ 1 },{ 3d }_{ yz }^{ 1 },{ 3d }_{ { xz } }^{ 1 }$

For which of the following elements all of its existing ion,

$M^{x+}$ , will be diamagnetic?

0%
$Cu$
0%
$Fe$
0%
$Cr$
0%
$Na$

Explanation

A neutral sodium atom is likely to achieve an octet in its outermost shell by losing its one valence electron. The cation produced in this way is Na $^+$ , having electronic configuration [Ne] 3s $^0$ which means there is no unpaired electron.

Hence, Na is diamagnetic in its existing ion form.

A transition metal

$A$ has spin only magnetic moment value of

$1.8\ B.M$ . When it is reacted with dilute

$H_2SO_4$ in presence of air, its compound

$B$ reacts with compound

$C$ to give compound

$D$ with liberation of iodine. Then the metal

$A$ , and compounds

$B, C$ and

$D$ are respectively:

0%
$Ti, TiSO_4, KI$ and $TiI_2$
0%
$An, ZnSO_4, KI$ and $ZnI_2$
0%
$Cu, CuSO_4, KI$ and $Cu_2I_2$
0%
$Cu, CuSO_4, KI$ and $CuI_2$

Match the catalysts (Column I) with products column (II)

Column I	Column II
(A) ${V}_{2}{O}_{5}$	(i) Polyethylene
(B) $Ti{Cl}_{4}/Al{(Me)}_{3}$	(ii) ethanal
(C) $Pd{Cl}_{2}$	(iii) ${H}_{2}{SO}_{4}$
(D) Iron Oxide	(iv) ${NH}_{3}$

0%
(A)- (ii); (B)- (iii); (C)- (i); (D)- (iv)
0%
(A)- (iii); (B)- (i); (C)- (ii); (D)- (iv)
0%
(A)- (iii); (B)- (iv); (C)- (i); (D)- (ii)
0%
(A)- (iv); (B)- (iii); (C)- (ii); (D)- (i)

Explanation

$V_2O_5$ is used as a catalyst in contact process for

$H_2SO_4$

$$TiCl_4/Al(Me)_3 is used as a catalyst in Ziegler Natta salt used as the catalyst for polymerisation of ethane to produce Polyethylene

$PdCl_2 \rightarrow$ used as catalyst for ethanal (Wacker process)

Iron oxide

$\rightarrow$ is used as a catalyst in Haber's synthesis of ammonia.

Hence, the correct option is

$D$

Match the following catalyst with their products-

Sr	Column 1	Sr	Column 2
1.	$TiCl_4+AlCl_3$	a.	Ethanal
2.	$V_2O_5$	b.	$NH_3$
3.	$Fe$	c.	Polyethylene
4.	$Pd$	d.	$H_2SO_4$

0%
$(i)-B ; (ii)-A ; (iii)-D ; (iv)-C$
0%
$(i)-C ; (ii)-D ; (iii)-B ; (iv)-A$
0%
$(i)-D ; (ii)-A ; (iii)-B; (iv)-C$
0%
$(i)-A ; (ii)-B ; (iii)-C ; (iv)-D$

Explanation

Solution:- (B)

$\left( i \right) - C; \; \left( ii \right) - D; \; \left( iii \right) - B; \; \left( iv \right) - A$

$\left( 1 \right) Ti{Cl}_{4} + Al{Cl}_{3} \rightarrow \text{Polyethylene} \left( c \right)$

$\left( 2 \right) {V}_{2}{O}_{5} \rightarrow {H}_{2}S{O}_{4} \left( d \right)$

$\left( 3 \right) Fe \rightarrow N{H}_{3} \left( b \right)$

$\left( 4 \right) Pd \rightarrow \text{Ethanal} \left( a \right)$

Consider the hydrates ions of

$Ti^{2+}, V^{2+}, Ti^{3+}$ and

$Sc^{3+}$ . The correct order of their spin-only magnetic moments is:

0%
$Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$
0%
$Ti^{3+} < Ti^{2+} < Sc^{3+} < V^{2+}$
0%
$Sc^{3+} < Ti^{3+} < V^{2+} < Ti^{2+}$
0%
$V^{2+} < Ti^{2+} < Ti^{3+} < Sc^{3+}$

Explanation

Solution:- (A)

$Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$

$a)\ Ti^{2+} = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2$

Unpaired electrons

$=2$

Spin only magnetic moment

$(\mu) = \sqrt{2(2+2)}$

$=\sqrt{8} B.M$

$b)\ Ti^{3+} = 1s^22s^22p^6 3s^2 3p^6 3d^1$

Unpaired electrons

$=1$

$\mu =\sqrt{1(1+2)} = \sqrt{3} B.M$

$c)\ V^{2+} = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$

$\mu = \sqrt{3(3+2)} = \sqrt{15} B.M$

$d)\ Sc^{3+} = 1s^2 2s^2 2p^6 3s^2 3p^6$

$\mu = 0$

What is the value of spin only magnetic moment of anionic and cationic part of complex

${\left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }_{2}$

$\left[ Fe{ \left( CN \right) }_{ 6 } \right]$ ?

0%
$4.9$ B.M and Zero
0%
Zero and $4.9$ B.M
0%
$2.9$ B.M and $0$
0%
$0$ and $2.9$ B.M

Explanation

${ \left[ Fe{ \left( CN \right) }_{ 6 } \right] }^{ -4 }$

${ Fe }^{ +2 }\longrightarrow 3{ d }^{ 6 }\quad$

${CN}^{-}\rightarrow$ S.F.L

${t}_{2g}^{2,2,2}$

${eg}^{0,0}$

No. of unpaired

${e}^{-}=0$

$B.M=0$

${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +2 }$

${ Fe }^{ +2 }\longrightarrow 3{ d }^{ 6 }\quad$

${H}_{2}O\rightarrow$ W.FL

${t}_{2g}^{2,1,1}$

${eg}^{1,1}$

No. of unpaired electrons

${e}^{-}=4$

$\quad \mu =\sqrt { 4\times 6 } =\sqrt { 24 } B.M=4.9B.M$

Photochromic glass, used for sunglasses, darkens when exposed to bright light and becomes
more transparent again when the light is less bright. The darkness of the glass is related to the
concentration of silver atoms.
The following reactions are involved.
reaction 1

$Ag^{+} + Cl^{-}\rightleftharpoons Ag + Cl$
reaction 2

$Cu^{+} + Cl \rightarrow Cu^{2+} + Cl^{-}$
reaction 3

$Cu^{2+} + Ag \rightarrow Cu^{+} + Ag^{+}$
Which statement about these reactions is correct?

0%
$Cu^{+}$ and $Cu^{2+}$ ions act as catalysts
0%
$Cu^{+}$ ions act as an oxidising agent in reaction $2$
0%
Reaction $3$ increases the darkness of the glass
0%
Silver atoms are reduced in reaction $3$

Explanation

Statement

$1$ is correct as the

${Cu}^{2+}$ and

${Cu}^{+}$ ions are regenerated at the end of reactions

$2$ and

$3$ respectively, hence they act as catalysts.

Statement

$2$ is incorrect as the oxidation number of

$Cl$ changes from

$0$ to

$-1$ , hence the

${Cu}^{+}$ ions act as a reducing agent.

Statement

$3$ is incorrect as silver atoms are converted to silver ions. Silver atoms increase the darkness of the glass whereas silver ions cause the glass to become more transparent.

Statement

$4$ is incorrect as the oxidation number of silver increases from

$0$ to

$+1$ , hence the silver atoms are oxidised.

The calculated spin-only magnetic moments (BM) of the anionic and cationic species of

$[Fe(H_2O)_6]_2$ and

$[Fe(CN)_6]$ , respectively, are:

0%
$4.9$ and $0$
0%
$2.84$ and $5.92$
0%
$0$ and $4.9$
0%
$0$ and $5.92$

Explanation

Solution:- (C)

$0$ and

$4.9$

Complex is

$[Fe(H_2O)_6]_2 [Fe(CN)_6]$

Complex ion	Configuration	No. of unpaired electrons	Magnetic moment
$[Fe(H_2O)_6]^{2+}$ $[Fe(CN)_6]^{4-}$	$t_{2g}^4 \,e_g^2$ $t_{2g}^6 \,e_g^0$	$4$ $0$	$4.9 \,BM$ $0$

The INCORRECT statement is:

0%
The spin - only magnetic moments of $[Fe(H_{2}O)_{6}]^{2+}$ and $[Cr(H_{2}O)_{6}]^{2+}$ are nearly similar
0%
The spin-only magnetic moment of $[Ni(NH_{3})_{4}(H_{2}O)_{2}]^{2+}$ is $2.83\ BM$
0%
The gemstone, ruby, has $Cr^{3+}$ ions occupying the octahedral sites of beryl
0%
The color of $[CoCl (NH_{3})_{5}]^{2+}$ is violet as it absorbs the yellow light

Explanation

(1)

$[Fe(H_{2}O)_{6}]^{2+}, Fe^{2+} \rightarrow 3d^{6} \rightarrow 4$ unpaired electron

$[Cr(H_{2}O)_{6}]^{2+}, Cr^{2+}\rightarrow 3d^{4}\rightarrow 4$ unpaired electron

(2)

$[Ni(NH_{3})_{4}(H_{2}O)_{2}]^{2+} ,\ Ni^{2+}\rightarrow 3d^{8}\rightarrow 2$ unpaired electron

$\mu_{m} = 2.83\ B.M$

(3) In gemstone, ruby has

$Cr^{3+}$ ion occupying the octahedral sites of aluminium oxide

$(Al_{2}O_{3})$ normally occupied by

$Al^{3+}$ ion.

(4) The complementary color of violet is yellow.

Which of the following complex is paramagnetic?

0%
$K_2[Ni(CN)_4]$
0%
$K_4[Fe(CN)_6]$
0%
$[Co(NH_3)_6]Cl_3$
0%
$[Cr(H_2O)_6]SO_4$

Explanation

The electronic configurations of

$Ni, Fe, Co$ and

$Cr$ in the given complexes are

$(3d)^8, (3d)^6, (3d)^6$ and

$(3d)^4$ , respectively.

These involve

$dsp^2, d^2sp^3, d^2sp^3$ and

$sp^3d^2$ -hybridization, respectively.

The complex

$[Cr(H_2O)_6]SO_4$ will be paramagnetic as Cr contains four unpaired electrons.

Which of the pair of complex compounds are tetrahedral as well as diamagnetic?

0%
$[CoCl_4]^-$ and $[Co(CO)_4]^-$
0%
$[Ag(SCN)_4]^{2-}$ and $[NiCl_4]^{2-}$
0%
$[Co(CO)_4]^-$ and $[Ni(CN)_4]^{4-}$
0%
$[PdCl_4]^{2-}$ and $[Ni(CN)_4]^{2-}$

The magnetic moment of a certain complex(A) of Co was found to be

$4.89$ B.M. and the EAN as

$36$ . Co also forms complex(B) with magnetic moment

$3.87$ B.M. and EAN

$37$ , and complex(C) with EAN as

$36$ but diamagnetic. Which of the following statements is true regarding the above observation?

0%
The oxidation states of Co in (A), (B) and (C) are $+3, +2$ and $+3$ , respectively
0%
Complexes (A) and (B) have $sp^3d^2$ hybridization state while (C) has $dsp^3$ hybridizaiton state
0%
The spin multiplicities of Co in (A), (B) and (C) are $4, 3$ and $1$ , respectively
0%
The oxidation states of Co in (A), (B) and (C) are $+6, +8$ and$ +1$$, respectively

Which of the following complexes has magnetic moment of

$2.83$ Bohr magneton?

0%
$[Ni(NH_3)_6]^{2+}$
0%
$[Ni(CN)_4]^{2-}$
0%
$TiCl_4$
0%
$[CoCl_6]^{3-}$

Explanation

$2.83$ Bohr magneton implies two unpaired electrons according to the expression

$\sqrt{n(n+2)}$ B.M.

The species

$Cu, Ni, Ti$ and

$Co$ in the given complexes have

$(3d)^8, (3d)^8, (3d)^0$ , and

$(3d)^6$ electronic configurations, respectively.

These involve

$sp^3 d^2, dsp^2, sp^3$ and

$sp^3d^2$ hybridization, respectively.

Thus, the complex

$[Ni(NH_3)_6]^{2+}$ has two unpaired electrons.

In the hydrogenenation of oil :-

0%
$Ni$ acts as a promoter
0%
$Cu$ acts as a catalyst
0%
Tellurium acts as a promoter
0%
$Ni$ acts as a catalyst

Among the following ions which one has the highest paramagnesium?

0%
$[Cr(H_2O)_6]^{3+}$
0%
$[Fe(H_2O)_6]^{2+}$
0%
$[Cu(H_2O)_6]^{2+}$
0%
$[Zn(H_2O)_6[^{2+}$

Which of the following anion exists in equilibrium condition with other anion depending upon the pH of the solution?

0%
$CrO^{2-}_{4}$
0%
$SO^{2-}_{4}$
0%
$CO^{2-}_{3}$
0%
$SO^{2-}_{3}$

Which of the following is incorrect about Wilkinson's catalyst?

0%
It is a non-ionic complex
0%
It is a diamagnetic complex
0%
It is a tetrahedral complex
0%
It is very effective for selective hydrogenation of organic molecule at room temperature and pressure

Which of the following complex(s) is/are paramagnetic in nature?

0%
$K_2[NiF_6]$
0%
$K_3[CoF_6]$
0%
$K_3[Fe(CN)_6]$
0%
$[Co(H_2O)_6]^{3+}$

Which of the following complex ion(s) are paramagnetic?

0%
$[Ni(CN)_4]^{2-}$
0%
$[NiCl_4]^{2-}$
0%
$[CoF_6]^{3-}$
0%
$[Co(NH_3)_6]^{3+}$

Select the paramagnetic complex compound.

0%
$K_4[Fe(CN)_5O_2]$
0%
$Fe(CO)_5$
0%
$[Cr(NH_3)_6]^{3+}$
0%
$[Ni(CN)_6]^{4-}$

Which of the following statement(s) is/are correct?

0%
$Ni(CO)_4$ -Tetrahedral, paramagnetic
0%
$[Ni(CN)_4]^{2-}$ -Square planar, diamagnetic
0%
$[Ni(CO)_4]$ -Tetrahedral, diamagnetic
0%
$[NiCl_4]^{2-}$ -Tetrahedral, paramagnetic

Select the correct statement(s):

0%
The transition metal ion is more polarizing compared to the alkali and alkaline earth metal ions of comparable size for a particular oxidation state
0%
The solubility of silver halides in water runs as $AgF > AgCl > AgBr > AgI$
0%
$LiCl$ is soluble in organic solvents while $NaCl$ is insoluble in such solvents
0%
All are incorrect

The pair(s) of reagents that yield paramagnetic species is/are:

0%
Na and excess of $NH_3$
0%
K and excess of $O_2$
0%
Cu and dilute $HNO_3$
0%
$O_2$ and 2-ethylanthraquinol

Explanation

(A) Na and excess of

$NH_{3}$ yield solvated electrons which are unpaired electrons. Hence, the species is paramagnetic.

(B) K and excess of

$O_{2}$ forms superoxide

$KO_{2}$ which contains paramagnetic superoxide ion

$O_{2}^{-}$ with one unpaired electron.

$HNO_{3}$ to form cupric nitrate with 1 unpaired electron in

$Cu^{2+}$ and also in NO. This results in paramagnetism.

(D) Refer image.

Hence , option A ,B & C are correct .

Which of the following complexes are diamagnetic?

0%
$[AuCl_4]$
0%
$[Co(H_2O)_6]^{3+}$
0%
$[CoF_6]^{3-}$
0%
$[Co(CO)_4]^-$

$Cr_2O^{2-}_7 \rightarrow{H^+} +Cr^{3+}.\ Eq.wt.of Cr_2O_7^{2-}$ is

0%
mol.wt/6
0%
mol wt./3
0%
mol. wt/4
0%
mol. wt/1

In the form of dichromate, Cr(VI) is a strong oxidizing agent in acidic medium but Mo(VI) in

$MoO_3$ and W(VI) in

$WO_3$ are not because......

0%
Cr(VI) is more stable than Mo(VI) and W (VI)
0%
Mo(VI) and W(VI) are more stable than Cr(VI)
0%
Higher oxidation states of heavier members of group-6 of transition series are more stable
0%
Lower oxidation states of heavier members of group-6 of transition series are more stable.

Which of the following sulphide is yellow in colour?

0%
$CuS$
0%
$CdS$
0%
$ZnS$
0%
$CoS$

What are the species A and B in the following.

$CrO_3 + H_2O \rightarrow A \xrightarrow{OH^-} B$

0%
$H_2CrO_4 , H_2Cr_2O_7$
0%
$H_2Cr_2O_7 , Cr_2O_3$
0%
$CrO_4^{2-} , Cr_2O_7^{-2}$
0%
$H_2Cr_2O_7 , CrO_4^{2-}$

Explanation

$CrO_3$ dissolves in water to produce orange colored acidic solution (

$Cr_2O_7^{2-}+2H^+$ ).

$2CrO_3+H_2O\rightarrow H_2Cr_2O_7$

$Cr_2O_7^{2-}+2OH^-\rightarrow 2CrO_4^{2-}+H_2O$

Therefore, A =

$H_2Cr_2O_7$ , B =

$CrO_4^{2-}$

Hence, the correct option is (D).

Maximum magnetic moments of

$Cr (Z = 24), Mn^{+} (Z=25),$ and

$Fe^{2+} (Z=26),$ are x, y, and z. They are in order:

0%
x < y < z
0%
x = z <y
0%
z < x = y
0%
x = y = z

The magnetic moment is associated with its spin angular momentum and orbital angular momentum,spin only magnetic value of

$Cr^{3+}$ ion is.......

0%
2.87 B.M
0%
3.87 B.M
0%
3.47 B.M
0%
3.57 B.M

Which of the following is a false statement?

0%
$Cr^{2+} (g)$ ion has greater magnetic moment compared to $Co^{3+} (g)$
0%
The magnitude of ionization potential of iron anion (monoanion) would be equal to electron gain enthalpy of iron
0%
Lanthanoids contraction is cause of lower I.P. of Pb than Sn
0%
If successive ionization energy are 332,738,849 , 4080,4958 (in Kj/mol) Then this element can be of 15 th group

Comprehension - 2
Potassium permanganate is prepared from the mineral pyrolusire

$MnO_2$ . Its crystal have deep purple color.It acts as an oxidizing agent in the netural,alkaline as well as acidic medium.In acidic medium,it is used in volumetric analysis for the estimation for ferrous salts,oxalates, etc.The titartion are carried out in the presence of

$H_2SO_4$ . However before using it as a titrant,it is first standardized with standard oxalic acid solution or Mohr salt solution.In one of the experiments on 13.4 g of dry pure sodium oxalate (molar mass =

$134 g mol^{-1} )$
was disolved in 100 mL of distilled water and then 100 mL of distilled water and then 100 mL of

$2.M H_2 SO_4$ were added.The solution was cooled to

$25.30^oC$ .Now to this solution 0.1 M

$KMnO_4$ Solution was added till a very faints pink color persisted.
(6) The purple color of

$K_2 MnO_4$ is due to :

0%
Incomplete d-subshell
0%
Ionic nature of $KMnO_4$
0%
Charge transfer
0%
Resonance in $MnO^-_4$

The radius of

$La^{3+}$ (At . No of

$La =57$ ) is

$1.06 \mathring{A}$ which one of the following given values will be closest to the radius of

$Lu^{3+}$ (at no. of

$Lu = 71$ ) ?

0%
$1.40 \mathring{A}$
0%
$1.06 \mathring{A}$
0%
$0.85 \mathring{A}$
0%
$1.60 \mathring{A}$

Chloro compound of vandium has only spin magnetic moment of 1.73 BM. This vanadium chloride has the formula (at.no V = 23)

0%
$VCl_4$
0%
$VCl_3$
0%
$VCl_2$
0%
$VCl_5$

Calomel

$(Hg_2Cl_2)$ on reaction with ammonium hydroxide gives:

0%
$HgO$
0%
$Hg_2O$
0%
$NH_2-Hg-Hg-Cl$
0%
$Hg NH_2Cl$

Which of the following substance are green ?

0%
$Fe(BO_{2})_{2}$
0%
$Cu$
0%
$Cr(BO_{2})_{3}$
0%
$Co(BO_{2})_{2}$

Which among the following are diamagnetic?

0%
$Cu^{2+}$
0%
$Zn^{2+}$
0%
$Ag^{+}$
0%
$Cd^{2+}$

General electronic configuration of actinodes is

$(n-2)f^{1-14} (n-1)d^{0-2} ns^2$ . Which of the following actinoids have one electrons in

$6d$ orbitals?

0%
$U$ (atomic no. $92$ )
0%
$Np$ (atomic no . $93$ )
0%
$Pu$ ( atomic no. $94$ )
0%
$Am$ ( Atomic no . $95$ )

Explanation

Actinoids are element 89 to 103 and fill their $5 \mathrm{f}$ shell. They are typical metals and radioactive also. $E g-T h, U, P u$ .

$U(z=92) \rightarrow 5 f^{3} 6 d^{1} 7 s^{2}$

$N{p}(z=93) \rightarrow 5 f^{4} 6 d^{1} 7 s^{2}$

$P u(z=94) \rightarrow 5 f^{6} 6 d^{0} 7 s^{2}$

$A m(z=95) \rightarrow 5 f^{7} 6 d^{\circ} 7 s^{2}$

So, U and $Np$ have one electron in $6 d$ orbital.

$(NH_4)_2 Cr_2 O_7$ on heating gives a gas which is also given by:

0%
Heating $NH_4NO_2$
0%
Heating $NH_4NO_3$
0%
$Mg_3N_2 + H_2O$
0%
Na(comp) + $H_2O_2$

On hearting ammonium dichromate the gas enolved is:

0%
Oxygen
0%
Ammonia
0%
Nitrous oxide
0%
Nitrogen

Among

$Ni(CO)_4, [Ni(CN)_4]^{2-}$ , and

$[NiCl_4]^{2-}$ .

0%
$Ni(CO)_4$ and $[NiCl_4]^{2-}$ are diamagnetic and $[Ni(CN)_4]^{2-}$ is paramagnetic
0%
$[NiCl_4]^{2-}$ and $[Ni(CN)_4]^{2-}$ are diamagnetic and $Ni(CO)_4$ is paramagnetic
0%
$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic and $[NiCl_4]^{2-}$ is paramagnetic
0%
$Ni(CO)_4$ is diamagnetic, and $[NiCl_4]^{2-}$ and $[Ni(CN)_4]^{2-}$ are paramagnetic

Explanation

a).

$[NiCl_4]^{2-}$

The oxidation number of Ni is

$+2$

Atomic number

$=28$

$Ni=[Ar]3d^84s^2$

$Ni^{2+}=[Ar]3d^8$

Chlorido is a weak field ligand, no pairing.

$sp^3$ - Hybridization (tetrahedral)

There are two unpaired electrons, so the complex is paramagnetic.

Spin magnetic moment

$\mu =\sqrt{n(n+2)}BM$

$=\sqrt{2(2+2)}BM=\sqrt{8}BM$

b).

$[Ni(CN)_4]^{2-}$

The oxidation number of Ni is

$+2$ .

Atomic number

$=28$

$Ni=[Ar]3d^84s^2$

$Ni^{2+}=[Ar]3d^8$

Cyano is a strong field ligand, it causes pairing.

$dsp^2$ - Hybridization (square planar)

There are no unpaired electrons so, the complex is diamagnetic.

Spin magnetic moment

$=$ zero

c).

$[Ni(CO)_4]$

The oxidation number of Ni is

$0$ .

Atomic number

$=28$

$Ni=[Ar]3d^84s^2$

$Ni=[Ar]3d^{10} \to$ in the complex.

CO is a strong field ligand, so pairing occurs

$sp^3$ - Hybridization (tetrahedral)

There are no unpaired electrons so, the complex is diamagnetic.

Spin magnetic moment

$=$ zero

Among the following complexes

$(K-P)$

$K_3[Fe(CN)_6](K), [Co(NH_3)_6]Cl_3(L),$

$Na_3[Co(oxalate)_3](M), [Ni(H_2O)_6]Cl_2(N),$

$K_2[Pt(CN)_4](O)$ and

$[Zn(H_2O)_6](NO_3)_2(P)$

The diamagnetic are:

0%
K, L, M, N
0%
K, M, O, P
0%
L, M, O, P
0%
L, M, N, O

Match the complex ions gives in Column-I with the colors given in Column-II and assign the correct code.

Column-I(Complex ion)	Column-II(Color)
(a) $[Co(NH_3)_6]^{3+}$	(p) Violet
(b) $[Ti(H_2O)_6]^{3+}$	(q) Green
(c) $[Ni(H_2O)_6]^{2+}$	(r) Pale blue
(d) $[Ni(H_2O)_4(en)]^{2+}(aq)$	(s) Yellowish orange
	(t)Blue

0%
a(p) b(q) c(s) d(t)
0%
a(s) b(r) c(q) d(p)
0%
a(r) b(q) c(s) d(p)
0%
a(s) b(p) c(q) d(r)

The 'spin-only' magnetic moment [in units of Bohr magneton,

$(\mu_B)$ ] of

$Ni^{2+}$ in aqueous solution would be (atomic number of

$Ni=28$ ).

0%
$6$
0%
$1.73$
0%
$2.84$
0%
$4.90$

Explanation

$Ni=[Ar]3d^84s^2;\quad Ni^{2+}=[Ar]3d^8$

$\mu_B=\sqrt{n(n+2)}$

$Ni^{2+}$ , unpaired electron is

$2$ .

$\therefore \mu_B=\sqrt{2(2+2)}=\sqrt{8}$

$\mu_B=2.828$

$\mu_B\approx 2.84\ BM$ .

Arrange the following ions in their magnetic moment:
(i)

$V^{4+}$
(ii)

$Mn^{4+}$
(iii)

$Fe^{3+}$
(iv)

$Ni^{2+}$
[At. Nos.

$V=23, Mn=25, Fe=26, Ni=28$ ]

0%
$(ii)>(iii)>(i)>(iv)$
0%
$(iii)>(iv)>(ii)>(i)$
0%
$(iii)>(ii)>(iv)>(i)$
0%
$(i)<(iv)<(iii)<(ii)$

Explanation

Magnetic moment depends upon the number of unpaired electrons:

(i)

$V^{4+}\,(Z=23):[Ar]3d^1$

$1$ unpaired electron

(ii)

$Mn^{4+}\,(Z=25):[Ar]3d^3$

$3$ unpaired electron

(iii)

$Fe^{3+}\,(Z=26):[Ar]3d^5$

$5$ unpaired electron

(iv)

$Ni^{2+}\,(Z=28):[Ar]3d^8$

$2$ unpaired electron

Number of unpaired electrons are maximum in

$Fe^{3+}$ . Hence it shows maximum magnetic moment and

$V^{4+}$ shows least magnetic moment because it has minimum number of unpaired electrons.

The order of magnetic moment is as:

$Fe^{3+}>Mn^{4+}>Ni^{2+}>V^{4+}$

Hence correct answer is

$(iii)>(ii)>(iv)>(i)$

Which one of the following statements is correct?

0%
$[Cu(NH_3)_4]^{2+}$ is diamagnetic while $[Fe(CN)_6]^{4-}$ is paramagnetic
0%
$[Cu(NH_3)_4]^{2+}$ is paramagnetic while $[Fe(CN)_6]^{4-}$ is diamagnetic
0%
Both are paramagnetic
0%
Both are diamagnetic

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 13 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 13 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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