Chemistry Test

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Chemistry Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

___________ is the unit for the rate of a chemical reaction.

A
Mole\(^{-1} s\)

B
Mole\({ }^{-1} s ^{-1}\)

C
Mole \(1 s\)

D
Mole\(^{-1}\)

Solution

The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:

1. The Rate of Disappearance of Reactants: \(\frac{[-\Delta \text { Reactants] }}{\Delta \text { time }}\)

2. The Rate of Formation of Products: \(\frac{[\Delta \text { Products }]}{\Delta \text { time }}\)

Thus, the units for the rate is Molarity per Seconds \(\left(\frac{M}{s}\right)\). or Moles per litre per second \(\left(\right.\)moles \(\left.{ }^{-1} s ^{-1}\right)\).
Question 2
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Which the following is not isostructural with \(\mathrm{SiCl}_{4}\)?

A
\(\mathrm{PbO}_{4}^{3-}\)

B
\(\mathrm{NH}_{4}^{+}\)

C
\(\mathrm{SCl}_{4}\)

D
\(\mathrm{SO}_{4}^{2-}\)

Solution

\(\mathrm{SiCl}_{4}\) is tetrahedral as it has four chlorine atoms attached tetrahedrally where as \(\mathrm{SCl}_{4}\) has four chlorine atoms and a lone pair on sulfur which makes its structure see-saw and the remaining molecules are having tetrahedral geometry.

So, \(\mathrm{SCl}_{4}\) is not iso structural with \(\mathrm{SiCl}_{4}\).
Question 3
1 / -0

The I.U.P.A.C name of the coordination compound \(K _{3}\left[ Fe ( CN )_{6}\right]\) is:

A
Potassium hexacyanoferrate(II)

B
Potassium hexacyanoferrate(III)

C
Potassium hexacyanoiron(II)

D
Tripotassium hexacyanoiron(II)

Solution

The I.U.P.A.C name of the coordination compound \(K _{3}\left[ Fe ( CN )_{6}\right]\) is Potassium hexacyanoferrate(III).

In this complex, there are six CN so they are named hexacyano and the cyanide ligand has \(-1\) charge. So the iron overall has \(-3\) charge, so the name of the central metal atom ends with -ate followed by the charge in roman letters.

Thus, the name of the complex is Potassium hexacyanoferrate(III).
Question 4
1 / -0

The increasing order of basicity of the following compounds is:

A
(b) < (a) < (c) < (d)

B
(b) < (a) < (d) < (c)

C
(d) < (b) < (a) < (c)

D
(a) < (b) < (c) < (d)

Solution

Order of base nature depends on electron donation tendency.

In compound nitrogen is sp2 hybridized so least basic among all given compounds. the compound is very strong nitrogenous organic base as lone pair of one nitrogen delocalize in resonance and make another nitrogen negatively charged and conjugate acid have two equivalent resonating structure.

Thus it is most basic in given compounds.
Question 5
1 / -0

The electronic configuration of gadolinium (atomic number 64) is?

A
\([X e] 4 f^{3} 5 d^{5} 6 s^{2}\)

B

\([X e] 4 f^{7} 5 d^{2} 6 s^{1}\)

C
\(\left[X e] 4 f^{7} 5 d^{1} 6 s^{2}\right.\)

D
\([X e] 4 f^{8} 5 d^{6} 6 s^{2}\)

Solution

The electronic configuration of \(\mathrm{La}(\mathrm{Z}=57)\) is \([\mathrm{Xe}] 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\).

Therefore, further addition of electrons occurs in a lower energy 4 f-orbital till it is exactly half filled at \(\operatorname{Eu}(\mathrm{Z}=63)\).

Thus, the electronic configuration of Eu is \([\mathrm{Xe}] 4 \mathrm{f}^{7} 6 \mathrm{~s}^{2}\).

Therefore, addition of next electron doesn't occur in a more stable exactly halffilled \(4 \mathrm{f}^{7}\) shell but occur in a little higher energy \(5 \mathrm{~d}\)-orbital.

Thus, the electronic configuration of \(\mathrm{Gd}(\mathrm{Z}=64)\) is \([\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\).
Question 6
1 / -0

Which statement is correct?

A
Fumaric acid is stronger than Maleic acid

B
Maleic acid is stronger than fumaric acid

C
Both (A) and (B)

D
None of the above
Solution
Maleic acid is stronger than fumaric acid.
- Maleic acid is cis-butenedioic acid whereas fumaric acid is trans-butenedioic acid.
- Maleic acid is able to lose \(H^{+}\)ion and it results in the formation of intra hydrogen bond. Whereas fumaric acid being a trans isomer has strong interaction with both the oxygen atoms of each carboxylic group. Therefore, fumaric acid is unable to give hydrogen ions as compared to maleic acid.
- That is why, maleic acid is stronger than fumaric acid.
Question 7
1 / -0

The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is about:

A
0.70 m

B
0.50 m

C
0.90 m

D
0.80 m

Solution

As we know:

The relative lowering of the vapour pressure of an aqueous solution is equal to the mole fraction of solute.

According to Raoult's law:

\(X_{B}=\frac{m * M_{A}}{1000+m * M_{A}}\)

Here:

\(\mathrm{m}\) is the molality of the solute

M_\{A\} is the molecular mass.

\(X_{-}\{B\}\) is the mole fraction of the component \(B=0.0125\)

\(\begin{aligned}&X_{B}=\frac{m * M_{A}}{1000+m * M_{A}} \\&\Rightarrow 0.0125=\frac{m * 18}{1000+m * 18} \\&\Rightarrow 12.5+0.225 m=18 m \\&\Rightarrow 17.775 m=12.5 \\&\Rightarrow m=\frac{12.5}{17.775} \\&\Rightarrow 0.70 .\end{aligned}\)
Question 8
1 / -0

The stability of dihalides of \(\mathrm{Si}, \mathrm{Ge}, \mathrm{Sn}\) and \(\mathrm{Pb}\) increases steadily in the sequence:

A
\(\mathrm{PbX}_{2}<\mathrm{SnX}_{2}<\mathrm{GeX}_{2}<\mathrm{SiX}_{2}\)

B
\(\mathrm{GeX}_{2}<\mathrm{SiX}_{2}<\mathrm{SnX}_{2}<\mathrm{PbX}_{2}\)

C
\(\mathrm{SiX}_{2}<\mathrm{GeX}_{2}<\mathrm{PbX}_{2}<\mathrm{SnX}_{2}\)

D
\(\mathrm{SiX}_{2}<\mathrm{GeX}_{2}<\mathrm{SnX}_{2}<\mathrm{PbX}_{2}\)

Solution

Due to the inert pair effect (the reluctance ofns \({ }^{2}\) electrons of outermost shell to participate in bonding) the stability of \(\mathrm{M}^{2+}\) ions (of group IV elements) increases as we go down the group.

The stability of dihalides increases down the group:

\(\mathrm{SiX}_{2}<\mathrm{GeX}_{2}<\mathrm{PbX}_{2}<\mathrm{SnX}_{2}\)
Question 9
1 / -0

Which of the following terms are not associated with proteins?

A
Molecular transport

B
Enzymes

C
Sex hormones

D
Cellular Signaling and Molecular Transport

Solution

Sex hormones, also known as sex steroids, gonadocorticoids and gonadal steroids, are steroid hormones that interact with vertebrate steroid hormone receptors. The sex hormones include the androgens, estrogens, and progestogens. Their effects are mediated by slow genomic mechanisms through nuclear receptors as well as by fast nongenomic mechanisms through membrane-associated receptors and signaling cascades. The polypeptide hormones luteinizing hormone, follicle-stimulating hormone and gonadotropin-releasing hormone are usually not regarded as sex hormones, although they play major sex-related roles.
Question 10
1 / -0

Stephen's reduction converts ethane nitrile into:

A
Acetic Anhydride

B
Ethanal

C
Propanone

D
Ethyl carbylamine

Solution

Stephen's reduction converts ethane nitrile into ethanal or Acetaldehyde.

When ethane nitrile is treated with Sn/HCl, it undergoes Stephen's reaction to form Acetaldehyde.

The reaction is:

The reaction is more efficient when aromatic nitriles are used rather than aliphatic nitriles. The reaction is a redox reaction and electron-withdrawing substituents can improve the rate of the reaction.