cheMCQ

12:11

CSIR UGC NET Dec15 Q. 11

The order of increasing Brønsted acidity for boron hydrides is

A. B5H9 < B6H10 < B10H14

B. B10H14 < B5H9 < B6H10

C. B6H10 < B10H14 < B5H9

D. B10H14 < B6H10< B5H9

Soln : A. B5H9 < B6H10 < B10H14

Acidity of boron hydride depends on the size of borane greater the sizes higher will be the acidity. This is because the negative charge, formed upon deprotonation, can be better delocalized over a large anion with many boron atoms than over a small one.

Vishnu Mestry

CSIR-NET

12:01

CSIR UGC NET Dec15 Q.10

Correct statement for coulometry is

A. it is based on faraday’s law of electrolysis

B. it is a type of voltammetry

C. it is based on Ohm’s law

D. it uses ion selective electrode

Soln : A. it is based on faraday’s law of electrolysis

Coulometry is an analytical method in which we measure amount of amount of water transferred during an electrolysis reaction by measuring the amount of electricity consumed/produced.

Vishnu Mestry

CSIR-NET

11:38

CSIR UGC NET Dec15 Q.9

Spin motion of which of the following gives magnetic moment (a) Electron (b) Proton (c) Neutron

Correct answer is
A. a and b

B. b and c
C. a and c

D. a, b and c

Soln: D. a, b and c

Magnetic moment is induced by the spin of elementary particles (electron and quarks in the proton and neutron of atomic nuclei). * Neutron is electrically neutral but a non-zero magnetic moment because of its internal quark structure.

(Reference : P.W. Atkins, Quanta: A handbook of concepts. Oxford University)

Vishnu Mestry

CSIR-NET

01:23

CSIR UGC NET Dec15 Q.8

The structures of XeF₂ and XeO₂F₂ respectively are
A. bent, tetrahedral
B. linear, square planar
C. linear, see-saw
D. bent, see-saw

XeF₂ XeO₂F₂

Ans: C. linear, see-saw

The three 5p electrons are promoted to higher energy 5d sub-level. The 5s three 5p and one 5d orbitals hybridize to give five sp3d hybrid orbitals. The four singly occupied orbitals are used for bond formation to two fluorine and two oxygen atoms. The fifth hybrid orbital contains the lone pair. The other 5d electrons of xenon which do not take part in the hybridization scheme are involved in π-bond formation to two oxygen atoms

Vishnu Mestry

CSIR-NET

00:54

CSIR UGC NET Dec15 Q.7

Identify the complex ions in the sequential order when ferroin is used as an indicator in the titration of iron(II) with potassium dichromate. (phen = 1, 10- phenathroline)
A. [Fe(phen)3] 2+ and [Fe(phen)3] 3+
B. [Fe(phen)3] 3+ and [Fe(phen)3] 2+
C. [Fe(CN)6] 4- and [Fe(CN)6] 3-
D. [Fe(CN)6] 3- and [Fe(CN)6] 4-

Soln: A. [Fe(phen)3] 2+ and [Fe(phen)3] 3+

Expln:
Ferroin is the chemical compound with the formula [Fe(o-phen)₃]SO₄, where o-phen is an abbreviation for 1,10-phenanthroline, a bidentate ligand.

The most important application of dichromate is in its reaction with iron(II) in which it is often preferred to permanganate.
The relevant half reaction is :
Fe²⁺ → Fe³⁺ + e^- E° = -0.77 V
and the total reaction is:
Cr₂O₇^2- + 6 Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6 Fe³⁺ + 7H₂O E° = 0.56 V
Therefore, a sharp colour change from red to blue occurs on oxidation of the complex.
[Ph3Fe]2+ → [Ph3Fe]3+

The indicator is prepared by mixing equivalent quantities of iron(II) sulphate and 1, 10- phananthroline. The resulting Fe(II) complex sulphate is called Ferroin. The Fe(III) complex sulphate is called ferrin. The color of ferroin is so much more intense than that of ferrin.

(Reference: Harris, D. C. Quantitative Chemical Analysis (4th ed.). New York, NY: W. H. Freeman. ISBN 0-7167-2508-8.)
http://wwwchem.uwimona.edu.jm/lab_manuals/c10expt31.html

Vishnu Mestry

CSIR-NET

23:58

CSIR UGC NET Dec15 Q.6

The correct statement for Mn-O bond lengths in [Mn(H₂O)⁶]²⁺ is

A. All bonds are equal
B. Four bonds are longer than two others
C. Two bonds are longer than four others
D. They are shorter than the Mn-O bond in [MnO4] -

Vishnu Mestry