Chemistry Test

Result

Chemistry Test - 40

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 atomic numbers of vanadium (V) , chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23,24,25, and 26. Which one of these may be expected to have the highest second ionization enthalpy?

V

Mn

Fe

Cr

Solution

The second ionization potential values of Cu and Cr are sufficiently higher than those of neighbouring elements . This is because of the electronic configuration of Cu⁺ which is 3d¹⁰ (completely filled) and of Cr⁺ which is 3 d⁵ (half filled ),i.e for the second ionization potentials , the electron is to be removed from very stable configuration.

Concepts :
Main Concept :
Electronic configurations of d-block elementsThe general electronic configuration of transition elements is (n-1) d^1-10 ns^1-2. The (n-1) stands for inner shell and the d-orbitals may have one to ten electrons and the s-orbital of the outermost shell (n) may have one or two electrons. It is observed from the Figure that 4s orbital (l = 0 and n = 4) is of lower energy than 3d orbitals (l = 2 and n = 3) upto potassium (At. No.19). The energy of both these orbitals is almost same in case of calcium (At. No. 20), but the energy of 3d orbitals decreases with further increase of nuclear charge and becomes lower than 4s, and 4p, (in case of scandium At. No.21). Thus after filling of 4s orbital successively with two electrons at atomic number 19 and 20, the next incoming electron goes to 3d orbital instead of 4p, as the former is of lower energy than the latter. This means that 21st electron enters the underlying principal quantum level with n = 3 rather than the outermost level with n = 4 which started filling at potassium (At. No.19), the first element of the fourth period. In the case of next nine elements following calcium, the incoming electron is filled in the d-subshell. Since half filled and completely filled subshells are stabler than the one in which one electron is short, an electron gets transferred from 4s to 3d in case of the elements with atomic number 24 and 29. Consequently, configuration of chromium and copper have only one 4s electron.

Electronic configuration of first series( or 3d) transition elements

Element	Symbol	Z	Electronic Configuration
Scandium	Sc	21	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²
Titanium	Ti	22	1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Vanadium	V	23	1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²
Chromium	Cr	24	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Manganese	Mn	25	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
Iron	Fe	26	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Cobalt	Co	27	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ 4s²
Nickel	Ni	28	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
Copper	Cu	29	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
Zinc	Zn	30	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²

The general electronic configuration of transition elements is (n-1) d^1-10 ns^1-2. The (n-1) stands for inner shell and the d-orbitals may have one to ten electrons and the s-orbital of the outermost shell (n) may have one or two electrons. It is observed from the Figure that 4s orbital (l = 0 and n = 4) is of lower energy than 3d orbitals (l = 2 and n = 3) upto potassium (At. No.19). The energy of both these orbitals is almost same in case of calcium (At. No. 20), but the energy of 3d orbitals decreases with further increase of nuclear charge and becomes lower than 4s, and 4p, (in case of scandium At. No.21). Thus after filling of 4s orbital successively with two electrons at atomic number 19 and 20, the next incoming electron goes to 3d orbital instead of 4p, as the former is of lower energy than the latter. This means that 21st electron enters the underlying principal quantum level with n = 3 rather than the outermost level with n = 4 which started filling at potassium (At. No.19), the first element of the fourth period. In the case of next nine elements following calcium, the incoming electron is filled in the d-subshell. Since half filled and completely filled subshells are stabler than the one in which one electron is short, an electron gets transferred from 4s to 3d in case of the elements with atomic number 24 and 29. Consequently, configuration of chromium and copper have only one 4s electron.

Electronic configuration of first series( or 3d) transition elements

Element	Symbol	Z	Electronic Configuration
Scandium	Sc	21	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²
Titanium	Ti	22	1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Vanadium	V	23	1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²
Chromium	Cr	24	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Manganese	Mn	25	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
Iron	Fe	26	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Cobalt	Co	27	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ 4s²
Nickel	Ni	28	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
Copper	Cu	29	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
Zinc	Zn	30	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²

Ionisation energies of transition elementsIonisation energies : The ionisation energies of the elements of first transition series are are given below:

Elements	l₁	l₂	l₃
Sc	632	1245	2450
Ti	659	1320	2721
V	650	1376	2873
Cr	652	1635	2994
Mn	716	1513	3258
Fe	762	1563	2963
Co	758	1647	3237
Ni	736	1756	3400
Cu	744	1961	3560
Zn	906	1736	3838

* in kJ mol^-1

The following generalizations can be obtained from the ionisation energy values given above.

(i) The ionisation energies or these elements are high and in the most cases lie those of s- and p-block elernents. This indicates that the transition elements are less electropositive than s-block elements.

Explanation : Transition metals have smaller atomic radii and higher nuclear charge as compared to the alkali metals. Both these factors tend to increase the ionisation energy, as observed.

(ii) The ionisation energy in any transition series increases in the nuclear with atomic number; the increase however is not smooth and as sharp as seen in the case of s and p-block elernents.

Explanation : The ionisation energy increases due to the increase in the nuclear charge with atomic number at the beginning of the series. Gradually, the shielding effect of the added electrons also increases. This shielding effect tends to decrease the attraction due to the nuclear charge. These two opposing factors lead to a rather gradual increase in the ionisation energies in any transition series.

(iii)The first ionisation energies of 5d-series of elements are much higher than those of the 3d and 4d series elements.

Explanation : In the 5d-series of transitions elements, after lanthanum (La), the added electrons to the next inner 4f orbitals. The 4f electrons have shielding effect. As a result, the outermost electrons experience greater nuclear attraction. This leads to higher ionisation energies for the 5d- series of transition elements.Ionisation energies : The ionisation energies of the elements of first transition series are are given below:

Elements	l₁	l₂	l₃
Sc	632	1245	2450
Ti	659	1320	2721
V	650	1376	2873
Cr	652	1635	2994
Mn	716	1513	3258
Fe	762	1563	2963
Co	758	1647	3237
Ni	736	1756	3400
Cu	744	1961	3560
Zn	906	1736	3838

Hence, the correct option is (D)

Question 2
1 / -0

When MnO₂ is fused with KOH, a coloured compound is formed, the product and its colour is

A
KMnO₄ , purple

B
Mn₂ O₃ , brown

C
Mn₃ O₄, black

D
K₂ MnO₄ , purple green

Solution

$\mathrm{K}_{2} \mathrm{MnO}_{4}$ (purple green) is formed and this also is the first step of preparation of $\mathrm{KMnO}_{4}$.
$2 \mathrm{MnO}_{2}+4 \mathrm{KOH}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{K}_{2} \mathrm{MnO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$
purple green
Hence, the correct option is (D)
Question 3
1 / -0

Mixture $X=0.02$ mole of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}$ and 0.02 mole of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$

was prepared in 2 L of solution. 1 Lof mixture $X+\operatorname{excess} A g N O_{3} \rightarrow Y$

$1 \mathrm{L}$ of mixture $\mathrm{X}+\mathrm{excess} \mathrm{BaCl}_{2} \rightarrow \mathrm{Z}$

Number of moles of $\mathrm{Y}$ and $\mathrm{Z}$ are :

A
0.01,0.01

B
0.02, 0.01

C
0.01, 0.02

D
0.02, 0.02

Solution

In $1 \mathrm{L}$ solution, there will be 0.01 mole of each $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$. Addition of excess $\mathrm{AgNO}_{3}$ will give $0.01 \mathrm{mole}$ of $\mathrm{AgBr}$. Addition of
excess $\mathrm{BaCl}_{2}$ will give $0.01 \mathrm{mole}$ of $\mathrm{BaSO}_{4}$
Hence, the correct option is (A)
Question 4
1 / -0

6.2 g of a sample containing Na₂CO₃, NaHCO₃ and non - volatile inert impurity on gentle heating loses 5% of its weight due to reaction 2NaHCO₃ $⟶$ Na₂CO₃ + H₂O + CO₂. Residue is dissolved in water and formed 100 mL solution and its 10 mL portion requires 7.5 mL of 0.2 M aqueous solution of BaCl₂ for complete precipitation of carbonates.
Determine weight (in gram) of Na₂CO₃ in the original sample.

A
1.59

B
0.53

C
1.06

D
None of these

Solution

Loss in Mass $=5 \times 10^{-2} \times 6.2=31 \times 10^{-2} \mathrm{g}$
$\begin{array}{c}2 N a H C O_{3} \\ 168 g\end{array} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\underbrace{\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}}_{62 \mathrm{g}}$
$62 \mathrm{g}$ loss in mass corresponds to $168 \mathrm{g} \mathrm{NaHCO}_{3}$
$31 \times 10^{-2} \mathrm{g}$ loss in mass corresponds to $\frac{168}{62} \times 31 \times 10^{-2}$
$=84 \times 10^{-2} \mathrm{g}$
$=10^{-2}$ moles of $\mathrm{NaHCO}_{3}$
Moles of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ produced from $\mathrm{NaHCO}_{3}=5 \times 10^{-3} \mathrm{moles}=5 \mathrm{mM}$
$\mathrm{mM}$ of $\mathrm{BaCl}_{2}=10 \times 7.5 \times 0.2=15 \mathrm{mM}$
Total $\mathrm{Na}_{2} \mathrm{CO}_{3}=15 \mathrm{mM}$
Initial $\mathrm{Na}_{2} \mathrm{CO}_{3}=15 \mathrm{mM}-5 \mathrm{mM}=10 \mathrm{mM}$
$=10 \times 10^{-3} \times 106$
$=1.06 \mathrm{g}$
Hence, the correct option is (C)
Question 5
1 / -0

Hydrolysis of phenyl isocyanide forms :

A
Benzoic acid

B
Formic acid

C
Acetanilide

D
Acetic acid

Solution

Note. Isocyanidesare hydrolysed only by acids, and not by alkalies. It is because –ve charge present on the carbon atom in isocyanides initially attracts electrophile (i.e. H⁺)

Hence, the correct option is (B)
Question 6
1 / -0

Number of oxygen atoms shared per $\mathrm{SiO}_{4}^{4-}$tetrahedron in

(i) Two dimensional sheet structured silicates,

(ii) Cyclic silicates and

(iii) Single strand chain silicates are, respectively

A
4, 2, 2

B
4, 3, 2

C
3, 3, 2

D
3, 2, 2

Solution

In two dimensional sheet silicates, three oxygens of each $\mathrm{Si} \mathrm{O}_{4}^{4-}$ units are shared. Thus contain $\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)^{2-}$ type anions. Cyclic silicates are obtained by sharing of two oxygens of each $\mathrm{Si} \mathrm{O}_{4}^{4-}$ tetrahedron. Chain silicates are also formed by the sharing of two oxygen atoms of each $\mathrm{Si} \mathrm{O}_{4}^{4-}$ units.
KEY CONCEPTS
Structural formula of ring silicate Structural formula: $\left(\mathrm{SiO}_{3}\right)^{2-}$

Number of oxygen atom (s) shared : 2
Net charge : $\mathrm{Si}=+4$
$\frac{\mathrm{O}=-6}{\mathrm{Net}=-2}$
Example of minerals : Beryl, Be ${ }_{3} \mathrm{Al}_{2}\left(\mathrm{SiO}_{3}\right)_{6}$
Hence, the correct option is (D)
Question 7
1 / -0

What is the correct relationship between the pHs of isomolar solutions of sodium oxide (pH₁), sodium sulphide (pH₂), sodium selenide (pH₃) and sodium telluride (pH₄) ?

A
pH₁ > pH₂ > pH₃ > pH₄

B
pH₁ > pH₂ $≫$ pH₃ > pH₄

C
pH₁ < pH₂ < pH₃ < pH₄

D
pH₁ < pH₂ < pH₃ $≫$ pH₄

Solution

Na₂O basic character

Na₂S decreases down the group
Na₂Se

Na₂Te Means Na₂O is most basic

We know that more the basic compound more is the pH [pH $\propto$ basic character]

Hence pH₁ > pH₂ > pH₃ > pH₄
Concepts :
Main Concept :
Properties of sodium oxide (Na₂O)Sodium oxide is ionic, white amorphous solid and strongly basic.

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

Column I                      Column II
(A) He                          (i) High electron gain enthalpy
(B) Cl                         (ii) Most electropositive element
(C) Ca                          (iii) Strongest reducing agent
(D) Li                            (iv) Highest ionization energy

The correct match of contents in Column I with those in Column II is:

A
A------(iii), B-----(i), C------(ii), D------(iv)

B
A-----(iv), B------(iii), C------(ii), D------(i)

C
A-----(i), B-----(ii),   C-----(iii), D-----(iv)

D
A-----(iv), B------(i), C------(ii), D-----(iii)

Solution

He - Inert gas, Cl - Electron gain enthalpy highest Ca - most electropositive metal, Li - ( strong reducing agent ). He the electronic configuration of $1 \mathrm{s}^{2}$ very stable $\mathrm{Cl} \longrightarrow \mathrm{Cl}^{-}$ gives highest $\Delta_{\mathrm{eg}} \mathrm{H}$ as $\mathrm{C}^{-}$ has stable electronic configuration $_{18}[\mathrm{Ar}]$
$20 \mathrm{Ca} \underset{-2 \mathrm{e}}{\longrightarrow} \mathrm{Ca}^{2+}$ has stable electronic configuration of $_{18}[\mathrm{Ar}]$
$\mathrm{E}_{\mathrm{red}}^{\mathrm{o}} \mathrm{Li}$ is $-3.07 \mathrm{V}$ strong reducing agent
Hence, the correct option is (D)
Question 9
1 / -0

Consider the following reaction.

Identify the structure of the major product X.

A

B

C

D

Solution

is the tertiary free radical which is the most stable amongst the given options.
Hence, the correct option is (B)
Question 10
1 / -0

Which of the following is present in DNA ?

A
Deoxyribose

B
Starch

C
Riboflavin

D
None of these

Solution

The sugar present in DNA is 2-deoxyribose while that present in RNA is ribose.

Concepts :
Main Concept :
Concept of Amino Acids: Structure of RNA and DNA

The sequence of bases along the DNA and RNA chain establishes its primary structure which controls the specific properties of the nucleic acid. An RNA molecule is usually a single chain of ribose-containing nucleotides. DNA molecule is a long and highly complex, spirally twisted, double helix, ladder like structure. The two polynucleotide chains or strands are linked up by hydrogen bonding between the nitrogenous base molecules of their nucleotide monomers. Adenine (purine) always links with thymine (pyrimidine) with the help of two hydrogen bonds and guanine (purine) with cytosine (pyrimidine) with the help of three hydrogen bonds. Hence, the two strands extend in opposite directions, i.e., are antiparallel and complimentary. The following fundamental relationship exist.

Thymine combines only with deoxyribose sugar and uracil only with ribose sugar. Other bases can combine with either of the two sugars.

I. The sum of purines equals the sum of pyrimidines.

II. The molar proportion of adenine equals to that of thymine.

III. The molar proportion of guanine equals to that of cytosine.

Watson, Crick and Witkins were awarded Noble prize in 1962 for suggesting the structure of DNA

Hence, the correct option is (A)