Chemistry Test

Result

Chemistry Test - 45

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The ionization isomer of [Cr(H₂O)₄Cl(NO₂)]Cl is :

A
[Cr(H₂O)₄(O₂N)]Cl₂

B
[Cr(H₂O)₄CI₂](NO₂ )

C
[Cr(H₂O)₄Cl(ONO)]Cl

D
[Cr(H₂O)₄Cl₂(NO₂)]. H₂O

Solution

Ionisation isomers are the complexes that produce different ions in solution, i.e, they have ions interchanged inside and outside the coordination sphere.

Cr(H₂O)₄CI(NO₂)]Cl and [Cr(H₂O)₄Cl₂](NO₂) have different ions outside the coordinationsphere and they are isomers. Therefore, they are ionisation isomers.
Hence, the correct option is (B)
Question 2
1 / -0

In the following electron dot structure calculate the formal charge from left to right nitrogen atom ?

A
-1 ,-1 , +1,

B
-1 ,+1 , -1

C
+1 ,-1, -1

D
+1,-1 , +1

Solution
Question 3
1 / -0

Which of the following molecules would be expected to be planar?

(1) NH₃

(2) SF₄

(3) XeF₄

(4) ICl^-₄

A
Only 1 ,2 and 3 are correct

B
Only 2 and 3 are correct

C
Only 3 and 4 are correct

D
Only 2 and 4 are correct

Solution

Concepts : Main Concept : Valence Shell Electron Pair Repulsion theory [VSEPR] for Molecules

The basic concept of the theory was suggested by Sidgwick and Powell (1940). It provides a useful idea for predicting shapes and geometries of molecules. The concept tells that, the arrangement of bonds around the central atom depends upon the repulsion operating between electron pairs(bonded or non bonded) around the central atom. Gillespie and Nyholm developed this concept as valence shell electron pair repulsions (VSEPR) theory. The main postulates of the VSEPR theory are :(1) For polyatomic molecules containing 3 or more atoms, one of the atoms is called the central atom to which other atoms are linked.(2) The geometry of a molecule depends upon the total number of valence shell electron pairs (bonded or not bonded) present around the central atom and their repulsion due to relative sizes and shapes.(3) If the central atom is surrounded by bond pairs only, then it gives the symmetrical shape to the molecule.(4) If the central atom is surrounded by lone pairs (lp) as well as bond pairs (bp) of electrons then the molecule has a distorted geometry (unsymmetrical shape).(5) The relative order of repulsion between electron pairs is as follows : $\mathrm{lp}-\mathrm{lp}>\mathrm{lp}-\mathrm{bp}>\mathrm{bp}-\mathrm{bp}$A lone pair is concentrated around the central atom while a bond pair is pulled out between two bonded atoms. As such repulsion becomes greater when a lone pair is involved.

Lone pair-bond pair and their respective shapes table

Hence, the correct option is (C)
Question 4
1 / -0

Which of the following reagents can be used to distinguish a phenol and an alcohol?

A
Ammoniacal AgNO₃

B
Ammoniacal Cu₂Cl₂

C
Aqueous ferric chloride

D
Neutral ferric chloride

Solution

Neutral ferric chloride gives characteristic violet colouration with phenol while alcohol remains unaffected. The colour is due to the formation of a complex as shown below :

$6 C_{6} H_{5} O H+F e^{3+} \rightarrow\left[F e\left(O C_{6} H_{5}\right)_{6}\right]^{3-}+6 H^{+}$

neutral medium $) \quad$ voiletcolour

key concept

Tests of phenol with Ferric chloride (voilet colored $\mathrm{Fe}(\mathrm{OAr})_{3}$ ) form $)$ Test for Water-Soluble Phenols

$3 \mathrm{ArOH}+\mathrm{FeCl}_{3} \stackrel{\text { Pyridine }}{\longrightarrow} \quad \mathrm{Fe}(\mathrm{OAr})_{3}$

Colored complex

Standard: Phenol

Procedure (for water-soluble phenols)

The iron (III) chloride test for phenols is not completely reliable for acidic phenols, but can be administered by dissolving $15 \mathrm{mg}$ of the unknown compound in $0.5 \mathrm{mL}$ of water or water-alcohol mixture and add 1 to 2 drops of $1 \%$ aqueous iron (III) chloride solution.

Positive test: A red, blue, green, or purple color is a positive test.

Cleaning up: Since the quantity of material is extremely small, the test solution can be diluted with water and flushed down the drain.
Hence, the correct option is (D)
Question 5
1 / -0

A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is

A
Pb²⁺

B
Hg²⁺

C
Cu²⁺

D
Co²⁺

Solution

The metal ion is $\mathrm{Hg}^{2+}$.
$H g^{2+}+2 I^{-} \longrightarrow H g I_{2} \downarrow($ Scarlet red precipitate $)$
$H g I_{2}+2 I^{-} \longrightarrow\left[H g I_{4}\right]^{2-}$
$H g^{2+}+C o(S C N)_{2} \longrightarrow H g(S C N)_{2} \downarrow+CO^{2+}$
Hence, the correct option is (B)
Question 6
1 / -0

If the bond length of CO bond in carbon monoxide is 1.128 Å, then what is the value of CO bond length in Fe(CO)₅?

A
1.15 Å

B
1.128 Å

C
1.72 Å

D
1.118 Å

Solution

In CO, bond order = 3 ; In metal carbonyls like Fe(CO)₅ due to dπ- pπ back-bonding so bond order of CO decreases slightly, therefore, bond length increases slightly. only choice 1.15 Å satisfies this fact.

Concepts :
Main Concept :
Examples on Sigma & Pi backbonded organometallics

Structure and properties
The metal-carbon bond in organometallic compounds is generally of character intermediate between ionic and covalent. Primarily ionic metal-carbon bonds are encountered either when the metal is very electropositive (as in the case of Group 1 or Group 2 metals) or when the carbon-containing ligand exists as a stable carbanion. Carbanions can be stabilized by resonance (as in the case of the aromatic cyclopentadienyl anion) or by the presence of electron-withdrawing substituents (as in the case of the triphenylmethyl anion). Hence, the bonding in compounds like sodium acetylide and triphenylmethylpotassium is primarily ionic. On the other hand, the ionic character of metal-carbon bonds in the organometallic compounds of transition metals, post-transition metals, and metalloids tends to be intermediate, owing to the middle-of-the-road electronegativity of such metals.
Organometallic compounds with bonds that have characters in between ionic and covalent are very important in industry, as they are both relatively stable in solutions and relatively ionic to undergo reactions. Two important classes are organolithium and Grignard reagents. In certain organometallic compounds such as ferrocene or dibenzenechromium, the pi orbitals of the organic moiety ligate the metal. In metal carbonyl and metal alkenes, back bonding(pi bonding) of electron density from metal to ligand antibonding orbitals makes stronger synergestic bonds.
Hence, the correct option is (A)
Question 7
1 / -0

Which species exhibits a plane of symmetry?

A

B

C

D

Solution

Only this compound has planeof symmetry.

Concepts :
Main Concept :
Symmetry operations in isomerismPlane of Symmetry: Imaginary plane passing through the molecule and bisecting in such a way that one part is mirrorimage of the other

(2) Center of symmetry: A center of symmetry in a molecule is said to exist if a line is drawn from any atom or group to this point and then extended to an equal distance beyond this point, meets the identical atom or group.
Center of symmetry is usually present only in an even membered ring.

(3) Axis of symmetry: It is an imaginary axis through which moleculerotates about certain degree then it gives equivalent configuration with respect to originalconfiguration. It is denoted by Cn
Where $n = \frac{360}{θ}$

Hence, the correct option is (D)
Question 8
1 / -0

A
It will behave ionic.

B
It will behave covalent.

C
It will behave neither covalent nor ionic.

D
Nothing can be said conclusively as per given information.

Solution

Concepts :
Main Concept :
Crystal structure of ionic compoundsIonic compounds generally have more complicated structures than metals. This is probably because:
(1) There are now at least two kinds of particles in the lattice, generally of different sizes.
(2) The cations attract the anions, but like ions repel one another. The structure must balance both types of forces.
(3) Many ions (e.g. nitrate, carbonate, azide) are very non-spherical in shape. They will thus pack differently in different directions.
One simple ionic structure is:
Cesium Chloride
Cesium chloride crystallizes in a cubic lattice. The unit cell may be depicted as shown. (Cs⁺ is teal, Cl^- is gold).

Hence, the correct option is (A)
Question 9
1 / -0

The standard reduction potentials of $\mathrm{Cu}^{2} / \mathrm{Cu}$ and $\mathrm{Cu}^{2} / \mathrm{Cu}$ are $0.337 \mathrm{V}$ and $0.153 \mathrm{V}$ respectively. The standard electrode potential of $\mathrm{Cu}^{+} / \mathrm{Cu}$ half cell is :

A
0.184V

B
0.827V

C
0.521V

D
0.490V

Solution

$\mathrm{E}^{\circ}$ is an intensive property
$\mathbf{E}^{\mathbf{o}}, \mathbf{\Delta} \mathbf{G}^{\mathbf{o}}=-\mathbf{n} \mathbf{E}^{\mathbf{o}} \mathbf{F}$
$(i) C u^{2+}+2 e^{-} \rightarrow C u \quad \mathrm{E}^{0}=-0.337 \mathrm{V} \Delta G_{1}^{0}=-0.674 \mathrm{F}$
$(i i) C u^{2+}+e^{-} \rightarrow C u^{+1} \quad \mathrm{E}^{0}=-0.153 \mathrm{V} \Delta G_{2}^{0}=-0.153 \mathrm{F}$
$C u^{+}+e^{-} \rightarrow C u \quad \Delta G_{3}^{o}=-n E^{o} F$
$\Delta G_{3}=\Delta G_{1}-\Delta G_{2}$
$-n F E^{0}=-0.674 F+0.153 F$
$\Rightarrow \quad \mathrm{E}^{\circ}=0.521 \mathrm{V} \quad(\because \mathrm{n}=1)$
Concepts :
Main Concept :
Standard electrode reduction potentialStandard electrode reduction potential
The values that we have just quoted for the two cells are actually the standard electrode potentials of the Mg2+ / Mg and Cu2+ / Cu systems.
The emf measured when a metal / metal ion electrode is coupled to a hydrogen electrode under standard conditions is known as the standard electrode potential of that metal / metal ion combination.
By convention, the hydrogen electrode is always written as the left-hand electrode of the cell. That means that the sign of the voltage quoted always gives you the sign of the metal electrode.
Standard electrode potential is given the symbol Eo
Hence, the correct option is (C)
Question 10
1 / -0

The ionisation energy of solid NaCl is 180 kcal per mol.The dissolution of the solid in water in the form of ions is endothermic to the extent of 1 kcal per mol. If the solvation energies of Na⁺ and Cl^- ions are in the ratio 6 : 5, what is the enthalpy of hydration of sodium ion ?

A
-85.6 kcal/mol

B
-97.6 kcal/mol

C
82.6 kcal/mol

D
+100 kcal/mol

Solution

$\Delta H_{\text {solution }}=\Delta H_{\text {ionization }}+\Delta H_{\text {hydration }}$
$\therefore 1=180+\Delta H_{h}$
$\Delta H_{h}=-179 \mathrm{kcal} \mathrm{mol}^{-1}$
Let $x$ be total heat of hydration $\left(\Delta H_{h}\right)$ and total
$\Delta H_{h}=\Delta H_{h\left(\mathrm{Na}^{+}\right)}+\Delta H_{h\left(\mathrm{Cl}^{-}\right)}$
$=\frac{6}{11} x+\frac{5}{11} x=x$
Thus, $\Delta H_{n}\left(\mathrm{Na}^{+}\right)=\frac{6 \times-179}{11}=-97 \cdot 63$ kcal $\mathrm{mol}^{-1}$
Hence, the correct option is (B)