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.

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.

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)

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

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

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] -

CSIR UGC NET Dec15 Q.5

The W-W bond order in [W(η⁵ -C5H5)(μ- Cl)(CO)₂]² is 

A. three

B. Two

C. one

D. Zero

Ans: 

To calculate bond order
(a) Calculate TVE : a
(b) Calculate n x 18 : b [n = number of metal]
(c) Calculate 𝑏−𝑎/2 = [number of M-M bond]

[W(η⁵ -C5H5)(μ- Cl)(CO)₂]²

(a) TVE = 36 [12 + 10 + 6 + 8] = a
(b) n x 18 = 2 x 18 = 36 = b
(c) 𝑏−𝑎/2 = 36−36 2 = 0 
Ans - D. Zero

CSIR UGC NET Dec-15 (Q-4)

Using crystal field theory, identify from the following complex ions that shows same µ_eff (spin only) values:

(A) [CoF₆]^3- , (B) [IrCl₆]^3- , (C) [Fe(H2O)₆]²⁺

(a) A and B 
(b) B and C 
(c) A and C 
(d) A, B and C

Ans:
All the complex are d⁶ and contain weak field ligand but except Ir all form high spin complex 

*[4d and 5d metals never form high spin complex] Hence spin only magnetic moment will be same for Co(III) and Fe(II)

The magnitude of the paramagnetism is expressed as an effective magnetic moment, μeff.

Magnetochemistry is concerned with the magnetic properties of chemical compounds. Magnetic properties arise from the spin and orbital angular momentum of the electrons contained in a compound. Compounds are diamagnetic when they contain no unpaired electrons. Molecular compounds that contain one or more unpaired electrons areparamagnetic. The magnitude of the paramagnetism is expressed as an effective magnetic moment, μeff.

Reference : Figgis&Lewis, pp. 424, 432

CSIR-NET Dec-2015 Q3

The oxidizing power of [CrO4]^2- , [MnO4]^2- , and [FeO4]^2- follows the order
(1.) [CrO4]^2- < [MnO4]^2- < [FeO4]^2-
(2.) [FeO4]^2- < [MnO4]^2- < [CrO4]^2-
(3.) [MnO4]^2- < [FeO4]^2- < [CrO4]^2-
(4.) [CrO4]^2- < [FeO4]^2- < [MnO4]^2-

Soln (1) 

The more an element is electronegative, the higher will be is oxidizing power. More electronegative is Fe (26), followed by Mn (25) and then Cr (24), so the order for the oxidizing power of those elements : Fe>Mn>Cr. 
Since oxidation number is same (VI) in all three metals.
[CrO4]^2- < [MnO4]^2- < [FeO4]^2-
Reference https://www.physicsforums.com/threads/a-problem-about-oxdizing-power.827918/

CSIR-NET Dec-2015 Q2

Deoxy – Hemocyanin is
(1.) heme protein and paramagnetic
(2.) Colorless and diamagnetic
(3.) O2 transport and paramagnetic
(4.) Blue colored and diamagnetic

Soln (2)
colorless Cu(I) deoxygenated form = d¹⁰ configuration (diamagnetic)

the blue Cu(II) oxygenated form.

Reference-
Hemocyanins (also spelled haemocyanins and abbreviated Hc) are proteins that transport oxygen throughout the bodies of some invertebrate animals. These metalloproteins contain two copper atoms that reversibly bind a single oxygen molecule (O2). They are second only to hemoglobin in frequency of use as an oxygen transport molecule. Unlike the hemoglobin in red blood cells found in vertebrates, hemocyanins are not bound to blood cells but are instead suspended directly in the hemolymph. Oxygenation causes a color change between the colorless Cu(I) deoxygenated form and the blue Cu(II) oxygenated form.
Coates CJ, Nairn J (July 2014). "Diverse immune functions of hemocyanins". Dev. Comp. Immunol. 45 (1): 43–55.doi:10.1016/j.dci.2014.01.021. PMID 24486681.

CSIR-NET Dec-2015 Q1

The biological functions of the cytochrome P₄₅₀ and myoglobin are, respectively 
(1.) Oxidation of alkene and O₂ storage 
(2.) O₂ transport and O₂ storage 
(3.) O2 storage and electron carrier 
(4.) Electron carrier and O₂ transport 

Soln 
(1) Oxidation of alkene (aliphatic)
The most common reaction catalyzed by cytochromes P₄₅₀ is a monooxygenase reaction, e.g., insertion of one atom of oxygen into the aliphatic position of an organic substrate (RH) while the other oxygen atom is reduced to water:

RH + O₂ + NADPH + H+ → ROH + H2O + NADP+

(Retrieved from https://en.wikipedia.org/wiki/Cytochrome_P450)

(2) O_{2 storage}
Myoglobin (symbol Mb or MB) is an iron- and oxygen-binding protein found in the muscle tissue of vertebrates in general and in almost all mammals. It is related to hemoglobin, which is the iron- and oxygen-binding protein in blood, specifically in the red blood cells.
Nelson DL, Cox MM (2000). Lehninger Principles of Biochemistry (3rd ed.). New York: Worth Publishers. p. 206.ISBN 0-7167-6203-X. (Google books link is the 2008 edition)