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Published on: 01/09/2022
QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Chemistry Subject - Transition and Inner Transition Elements, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Chemistry Test
1.
Which one of the following is not correct?
La(OH)3 is less basic than Lu(OH)3
In lanthanoid series ionic radius of Ln3+ ions decreases
La is actually an element of transition metal series rather than lanthanide series
Atomic radii of Zr and Hf are same because of lanthanide contract
2.
The actinoid elements which show the highest oxidation state of +7 are _______.
Np, Pu, Am
U, Fm, Th
U, Th, Md
Es, No, Lr
3.
The most common oxidation state of actinoids is _______.
+2
+3
+4
+6
4.
Which of the following oxidation states is most common among the lanthanoids?
+4
+2
+5
+3
5.
Which of the following lanthanoid ions is diamagnetic?
Eu2+
Yb2+
Ce2+
Sm2+
6.
Which one of the following statements related to lanthanons is incorrect?
Europium shows +2 oxidation state
The basicity decreases as the ionic radius decreases from Pr to Lu.
All the lanthanons are much more reactive than aluminium
Ce4+ solutions are widely used as oxidising agents in volumetric analysis.
7.
The number of moles of acidified KMnO4 required to oxidize 1 mole of ferrous oxalate(FeC2O4) is _______.
5
3
0.6
1.5
8.
9.
Permanganate ion changes to ________ in acidic medium.
MnO42−
Mn2+
Mn3+
MnO2
10.
Which of the following statements is not true?
on passing H2S, through acidified K2Cr2O7 solution, a milky colour is observed
Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis
K2Cr2O7 solution in acidic medium is orange in colour
K2Cr2O7 solution becomes yellow on increasing the PH beyond 7
11.
12.
The correct order of increasing oxidizing power in the series _______.
VO2+ < Cr2O72- < MnO4-
Cr2O72- < VO2+ < MnO4-
Cr2O72- < MnO4- < VO2+
MnO4- < Cr2O72- < VO2+
13.
The catalytic behaviour of transition metals and their compounds is ascribed mainly due to _______.
their magnetic behaviour
their unfilled d orbitals
their ability to adopt variable oxidation states
their chemical reactivity
14.
The magnetic moment of Mn2+ ion is _______.
5.92BM
2.80BM
8.95BM
3.90BM
15.
Which one of the following ions has the same number of unpaired electrons as present in V3+?
Ti3+
Fe3+
Ni2+
Cr3+
16.
Among the transition metals of 3d series, the one that has highest negative \(\left( \frac { M^{ 2+ } }{ M } \right) \) standard electrode potential is _______.
Ti
Cu
Mn
Zn
17.
Which of the following d block element has half filled penultimate d sub shell as well as half filled valence sub shell?
Cr
Pd
Pt
none of these
18.
Sc (Z = 21) is a transition element but Zinc (z = 30) is not because _______.
both Sc3+ and Zn2+ ions are colourless and form white compounds
In case of Sc, 3d orbital are partially filled but in Zn these are completely filled
last electron as assumed to be added to 4s level in case of zinc
both Sc and Zn do not exhibit variable oxidation states
1.
(a)
La(OH)3 is less basic than Lu(OH)3
2.
(a)
Np, Pu, Am
3.
(b)
+3
4.
(d)
+3
5.
Yb2+ - 4f14 - no unpaired electrons - diamagnetic
6.
As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.
7.
+7 +2+3 2+ 3+ 4+
MnO4- + FeC2O4 ⟶ Mn2+ + Fe3+ + 2CO2
5e- acception 3e- release
5 moles of FeC2O4 \(\equiv\) 3 moles of KMnO4
1 mole of FeC2O4 \(\equiv\) (3/5) moles of KMnO4
1 mole of FeC2O4 \(\equiv\) 0.6 moles of KMnO4
8.
(c)
9.
MnO-4 + 8H+ + 5e- → Mn2+ + 4H2O
10.
(b)
Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis
11.
(b)
12.
+5 +6 +7
VO2+ < Cr2O72- < MnO4-
Greater the oxidation state, higher is the oxidising power.
13.
(c)
their ability to adopt variable oxidation states
14.
Mn2+ ⇒ 3d5 contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)
15.
(c)
Ni2+
16.
(a)
Ti
17.
Cr ⇒ [Ar]3d54s1
18.
(b)
In case of Sc, 3d orbital are partially filled but in Zn these are completely filled
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