Chemistry Test

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Chemistry Test - 30

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Weekly Quiz Competition

Question 1
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Heating mixture of Cu₂O and Cu₂S will give

A
Cu + SO₃

B
CuO + CuS

C
Cu₂SO₃

D
Cu + SO₂

Solution

$2 C u_{2} O+C u_{2} S \rightarrow 6 C u+S O_{2}$

This reaction takes place during bessemerization and is auto reduction process to give copper metal.

Hence correct answer is option D.
Question 2
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The following species having which hybridisation ?

$S b C l_{6}^{-}, S n C l_{6}^{2-}, X e F_{5}^{+}$ and $I O_{6}^{5-}$

A
sp³ d²

B
sp³

C
sp

D
sp³ d³

Solution

As per VSEPR Theory
\begin{aligned}
S b C l_{6}^{-} \Rightarrow & \mathrm{H}=\frac{1}{2}(\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}) \\
&=\frac{1}{2}(5+6+1)=6 \mathrm{Sp}^{3} \mathrm{d}^{2} \text { hybridisation octahedral } \\
\mathrm{SnCl}_{6}^{2-} \Rightarrow \mathrm{H} &=\frac{1}{2}(4+6+2)=6 \mathrm{sp}^{3} \mathrm{d}^{2} \text { hybridisation }
\end{aligned}
$\mathrm{XeF}_{6} \Rightarrow \mathrm{H}=\frac{1}{2}(8+6)=7$ it is an exception it is octahedral geometry i.e. distorted octahedral with lone pair delocalised over faces of octahedron. $I O_{6}^{5-} \Rightarrow \mathrm{H}_{5} 1 \mathrm{O}_{6}$ is $(\mathrm{OH})_{5}$ । $\mathrm{O}$ are octahedral units.
Hence, the correct option is (A)
Question 3
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The most stable resonating structure of following compound is

A

B

C

D

Solution

Complete octet & Extended conjugation
Concepts :
Main Concept :
Stability of resonance structuresExample
The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on either terminal oxygen can be delocalized by resonance through the terminal oxygens.
Benzene is an extremely stable molecule and it is accounted for its geometry and molecular orbital interaction, but most importantly it's due to its resonance structures. The delocalized electrons in the benzene ring make the molecule very stable and with its characteristics of a nucleophile, it will react with a strong electrophile only and after the first reactivity, the substituted benzene will depend on its resonance to direct the next position for the reaction to add a second substituent.
The next molecule, the Amide, is a very stable molecule that is present in most biological systems, mainly in proteins. By studies of NMR spectroscopy and X-Ray crystallography it is confirmed that the stability of the amide is due to resonance which through molecular orbital interaction creates almost a double bond between the Nitrogen and the carbon.
Example: Multiple Resonance of other Molecules

Molecules with more than one resonance form
Some structural resonance conformations are the major contributor or the dominant forms that the molecule exists. For example, if we look at the above rules for estimating the stability of a molecule, we see that for the third molecule the first and second forms are the major contributors for the overall stability of the molecule. The nitrogen is more electronegative than carbon so, it can handle the negative charge more than carbon. A carbon with a negative charge is the least favorable conformation for the molecule to exist, so the last resonance form contributes very little for the stability of the Ion.
Hybrid resonance

The different resonance forms that the molecule has helps and directs the reactivity to specific sites.
The Hybrid Resonance forms show the different Lewis structures with the electron been delocalized. This is very important for the reactivity of chloro-benzene because in the presence of an electrophile it will react and the formation of another bond will be directed and determine by resonance. The long pair of electrons delocalized in the aromatic substituted ring is where it can potentially form a new bond with an electrophile, as it is shown there are three possible places that reactivity can take place, the first to react will take place at the para position with respect to the chloro substituent and then to either ortho position.
Hence, the correct option is (D)
Question 4
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The common features among the species CN^– , CO and NO⁺ are

A
Bond order three and isoelectronic

B
Bond order three and weak field ligand

C
Bond order two and $π$ - e^- acceptors

D
Isoelectronic and weak field ligands

Solution

$\mathrm{CN}^{\prime}$ ( Total Electron
14) $B \cdot 0=\frac{10-4}{2}=3$
co ( Total Electron
14) $\quad B .0=\frac{10-4}{2}=3$
Not ( Total electrons
14) $B .0=\frac{10-4}{2}=3$
Hence, the correct option is (A)
Question 5
1 / -0

Among $N O_{2}$ ⁺ KO₂, Na₂O₂, and NaAlO₂, which is paramagnetic ?

A
Na₂ O₂ only

B
Na₂O₂ and NaAlO₂

C
$\mathrm{KO}_{2}$ and $\mathrm{NO}_{2}$

D
KO₂ only

Solution

$\mathrm{NO}_{2}^{+}(22$ Electrons $)$
$O_{2}^{-}=\sigma 1 s^{2}<\tilde{\sigma} 1 s^{2}<\sigma 2 s^{2}<\dot{\sigma} 2 s^{2}<\sigma 2 P_{z}^{2}<\pi 2 P_{y}^{2}=\pi 2 P_{x}^{2}<\tilde{\pi} 2 P_{y}^{2}=\pi 2 P_{x}^{1}$ It has unpaired
electron so paramagnetic $O_{2}^{2-}=\sigma 1 s^{2}<\dot{\sigma} 1 s^{2}<\sigma 2 s^{2}<\dot{\sigma} 2 s^{2}<\sigma 2 P_{z}^{2}<\pi 2 P_{y}^{2}=\pi 2 P_{x}^{2}<\hat{\pi} 2 P_{y}^{2}=\hat{\pi} 2 P_{x}^{2}$ It has no unpaired
electron so diamagnetic $\mathrm{AlO}_{2}^{-}(30$ Electrons ) Paired Electrons
Hence, the correct option is (D)
Question 6
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Based on following information determine the value of $x$ and $y$ $\Rightarrow\left\{\begin{array}{cccc}\left(\mathrm{CH}_{3}\right)_{\mathrm{x}} \mathrm{AlCl}_{\mathrm{y}} & \mathrm{x}\left(\mathrm{CH}_{4}\right)(\mathrm{g}) & + & \underbrace{\mathrm{yCl}+\mathrm{Al}^{3+}}_{\downarrow} \\ & & \mathrm{AgCl}(\mathrm{s}) \\ 0.643(\mathrm{g}) & 0.222 \mathrm{g} & 0.9969 \mathrm{g} \\ \mathrm{I} & \mathrm{II} & \mathrm{III}\end{array}\right.$
atomic masses of $\mathrm{Al}$, $\mathrm{Cl}$ and $\mathrm{Ag}$ are respectively 27,35.5,108 .

A
$x = 2, y = 1$

B
$x = 2, y = 2$

C
$x = 3, y = 4$

D
X=2 , Y=3

Solution

Cl on PoAC $\Rightarrow$
$\frac{0.643}{15 x+27+35 \cdot 5 y} \times y=\frac{0.996}{143 \cdot 5} \times 1$
$\mathrm{C}$ on $\mathrm{PoAC} \Rightarrow$
$\frac{0.643}{15 x+27+35-5 y} \times x=\frac{0-222}{16} \times 1$
by divide $\mathrm{B}$ with $\mathrm{A}$
$\frac{A}{B} \Rightarrow \frac{x}{y}=\frac{\frac{0.222}{16}}{\frac{0.506}{143.5}}=2$
take least ratio in case of salt
$x=2$
$\therefore \mathrm{y}=1$
Hence, the correct option is (A)
Question 7
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A sample of an ideal gas is expanded from original volume of $1 \mathrm{m}^{3}$ to twice its volume in a reversible process for which $\mathrm{p}=\alpha \mathrm{V}^{2} \quad\left(\alpha=5 \mathrm{atm} \mathrm{m}^{-6}\right)$
Calculate
(i) Work done on $P$ - V diagram
(ii) Calculate $\Delta \mathrm{S}_{\mathrm{m}}$ if $\mathrm{C}_{\mathrm{Vm}}=20 \mathrm{J} \mathrm{K}^{-1} \mathrm{m}^{-1}, \mathrm{R}=8 \mathrm{JK}^{-1} \mathrm{m}^{-1}$ and $\mathrm{In} 2=0.7$
The values of work done $\& \Delta S_{m}$ may be

A

B
(i) $\frac{35}{3} \times 10^{5} \mathrm{J}$ (ii) $-45.82 \mathrm{J} \mathrm{m}^{-1} \mathrm{K}^{-1}$

C
(i) $\frac{7}{3} \times 10^{5} \mathrm{J}$ (ii) $-15.27 \mathrm{J} \mathrm{m}^{-1} \mathrm{K}^{-1}$

D
(i) $\frac{14}{3} \times 10^{5} \mathrm{J}$ (ii) $-30.54 \mathrm{J} \mathrm{m}^{-1} \mathrm{K}^{-1}$

Solution
Question 8
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In a cell that utilizes the reaction $Z n(s)+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$ addition
of $\mathrm{H}_{2} \mathrm{SO}_{4}$ to cathode compartment, will:

A
Lower the E and shift equilibrium to the left

B
Lower the E and shift the equilibrium to the right

C
Increase the E and shift the equilibrium to the right

D
Increase the E and shift the equilibrium to the left

Solution

$Z n(s)+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$
$E=E^{0}-\frac{0.059}{2} \log \frac{\left[Z n^{+2}\right]}{\left[H^{+}\right]^{2}}$
So as we add sulphuric acid, conc. of hydrogen ion increases and equilibrium shifts to right. Also according to the equation emf of the cell will be increased by adding acid.
Hence, the correct option is (C)
Question 9
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A crystal made up of particles X, Y, and Z. X forms fcc packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. It all particles along one body diagonal are removed, then the formula of the crystal would be

A
XYZ₂

B
X₂YZ₂

C
X₈Y₄Z₅

D
X₅Y₄Z₈

Solution

In $\mathrm{FCC}$, unit cell number of atoms present $=4$
Now, $2 \times$ Number of octahedral voids $=$ number of tetrahedral voids.
And number of octahedral voids $=$ Number of atoms present in cubic unit cell. So, $X=4, Y=4, Z=8$
On the body diagonal there is one octahedral void (at body center) and two tetrahedral voids (at $\frac{1}{4}$ of the distance from each corner). So, after removal of atoms on body diagonal, we have:$X=6 \times \frac{1}{2}+\frac{\theta-2}{8}=3+\frac{3}{4}=\frac{15}{4}$
(face) (corners) $Y=4-1=3 ; Z=8-2=6$
$\begin{array}{lllllll}\text { So, } & X & Y & Z & \Rightarrow & X & Y & Z\end{array}$
$\frac{15}{4} \times 43 \times 4 \quad 6 \times 4 \quad \frac{15}{5} \frac{12}{4} \frac{24}{8}$
So, formula of crystal is $X_{5} Y_{4} Z_{8}$
Hence, the correct option is (D)
Question 10
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A solution which is 10^–3 M each in Mn²⁺ , Fe²⁺ , Zn²⁺ and Hg²⁺ is treated with 10^{- 16} M sulphide ion. If K_spof MnS, FeS, ZnS and HgS are 10 ^{- 13} , 10^–18 , 10^–24 and 10^–53 respectively, which one will precipitate first ?

A
FeS

B
MgS

C
HgS

D
ZnS

Solution

$\left[\mathrm{S}^{2} \cdot\right]$ need for precipitation of
$\mathrm{FeS}=\frac{10^{-18}}{10^{-3}}=10^{-15} \mathrm{M}$
$\mathrm{MnS}=\frac{10^{-13}}{10^{-3}}=10^{-10} \mathrm{M}$
$\mathrm{HgS}=\frac{10^{-53}}{10^{3}}=10^{-50} \mathrm{M}$
$\mathrm{ZnS}=\frac{10^{-34}}{10^{-3}}=10^{-21} \mathrm{M}$
Thus minimum $\left[\mathrm{S}^{2} \cdot\right]$ for precipitation is for HgS it will be precipitated first
Hence, the correct option is (C)