Important Question paper

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Questions:

Multiple Choice Question

15 x 1 = 15

The characteristic feature of orderly arrangement of molecules belongs to ______________.

(a)

Solids

(b)

Liquid

(c)

Gases

(d)

None of these

Identify the incorrect statement about a compound.

(a)

A molecule cannot be separated into its constituent elements by physical methods of separation

(b)

A molecule of a compound has atoms of different elements

(c)

A compound retains the physical properties of its constituent element

(d)

The ratio of atoms of different elements in a compound is fixed

Consider the following statements
i) Matter possesses mass.
ii) 22-carat gold is a mixture.
iii) Dry ice is a compound.
Which of the following statement(s) given above is/ are correct?

(a)

1 & 3

(b)

Only 1

(c)

1 & 2

(d)

1,2, & 3

Two 22.4 litre containers A and B contains 8 g of O₂ and 8 g of SO₂ respectively at 273 K and 1 atm pressure, then ______.

(a)

Number of molecules in A and B are same

(b)

Number of molecules in B is more than that in A.

(c)

The ratio between the number of molecules in A= to number of molecules in B is 2:1

(d)

Number of molecules in B is three times greater than the number of molecules in A

Match the list-I with list-II and select the correct answer using the code given below the lists.

List-I		List-II
A	Cr₂O₇^2-	1	+5
B	MnO₄^-	2	+6
C	VO₃^-	3	+3
D	FeF₆³⁺	4	+7

(a)

A	B	C	D
3	1	4	2

(b)

A	B	C	D
4	3	2	1

(c)

A	B	C	D
2	4	1	3

(d)

A	B	C	D
3	2	1	4

Which of the following statement(s) is/are not true about the following decomposition reaction.
2KClO₃ \(\longrightarrow\) 2KCl + 3O₂
(i) Potassium is undergoing oxidation
(ii) Chlorine is undergoing oxidation
(iii) Oxygen is reduced
(iv) None of the species are undergoing oxidation and reduction.

(a)

only (iv)

(b)

(i) and (iv)

(c)

(iv) and (iii)

(d)

All of these

Identify disproportionation reaction

(a)

CH₄ + 2O₂ \(\longrightarrow\) CO₂+ 2H₂O

(b)

CH₄ + 4Cl₂ \(\longrightarrow\) CCl₄ + 4HCI

(c)

2F₂+ 2OH \(\longrightarrow\) 2F^-+ OF₂+ H₂O

(d)

2NO₂ + 2OH^- \(\longrightarrow\) NO^-₂ + NO^-₃ + H₂O

The oxidation number of Cr in Cr₂O₇^2- _______ is

(a)

(b)

-6

(c)

(d)

-7

Total number of electrons present in 1.7 g of ammonia is _____.

(a)

6.022\(\times\)10²³

(b)

\(\frac { 6.022\times { 10 }^{ 22 } }{ 1.7 } \quad \)

(c)

\(\frac { 6.022\times { 10 }^{ 24 } }{ 1.7 } \)

(d)

\(\frac { 6.022\times { 10 }^{ 23 } }{ 1.7 } \)

10.

The oxidation number of hydrogen in LiH is _________

(a)

(b)

-1

(c)

(d)

-2

11.

The oxidation number of oxygen in O₂ is__________

(a)

(b)

(c)

(d)

-2

12.

When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to ______.

(a)

2 moles of HCl (g)

(b)

0.5 moles of HCl (g)

(c)

1.5 moles of HCl (g)

(d)

1 moles of HCl (g)

13.

When 6.3 g of sodium bicarbonate is added to 30 g of the acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is _____.

(a)

(b)

0.75

(c)

0.075

(d)

0.3

14.

1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is ______________.

(a)

(b)

4.4%

(c)

16%

(d)

8.4%

15.

The number of water molecules in a drop of water weighing 0.018 g is ________.

(a)

6.022\(\times\)10²⁶

(b)

6.022\(\times\)10²³

(c)

6.022\(\times\)10²⁰

(d)

9.9 \(\times\)10²²

Generic 3

7 x 3 = 21

16.

Calculate the Formula Weights of the following compounds. NO₂

17.

How much mass (in gram units) is represented by the following?
5.14 mol of H₅IO₆

18.

How much mass (in gram units) is represented by the following ?
3.0 mol of CO₂

19.

How much mass (in gram units) is represented by the following ?
0.2 mol of NH₃

20.

One million silver atoms weigh 1.79 x 10^-16 g. Calculate the atomic mass of silver.

21.

Which contains the greatest number of moles of oxygen atoms
i) 1 mol of ethanol
ii) 1 mol of formic acid
iii) 1 mol of H₂O

22.

The density of carbon dioxide is equal to 1.965 kgm^-3 at 273 K and 1 atm pressure. calculate the molar mass of CO₂.

Generic 2

3 x 2 = 6

23.

How many orbitals are possible for n = 4?

24.

The stabilisation of a half filled d - orbital is more pronounced than that of the p-orbital why?

25.

Calculate the average atomic mass of naturally occurring magnesium using the following data.

Isotope	Isotopic atomic mass	Abundance(%)
Mg²⁴	23.99	78.99
Mg²⁵	24.99	10.00
Mg²⁶	25.98	11.01

Generic 5

5 x 5 = 25

26.

Balance the following equations by ion electron method.
\({ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }+{ I }_{ 2 }\longrightarrow { Na }_{ 2 }{ S }_{ 4 }{ O }_{ 6 }+NaI\)

27.

Balance the following equations by ion electron method
\({ C }_{ 2 }{ O }_{ 4 }^{ 2- }+{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { Cr }^{ 3+ }+{ CO }_{ 2 }\) (in acid medium)

28.

Balance the following equations by ion electron method.
i) \({ KMn }O_{ 4 }+{ SnCl }_{ 2 }+HCI\longrightarrow MnCI_{ 2 }+{ SnCI }_{ 4 }+{ H }_{ 2 }O+KCI\)
ii)
iii)
iv)

29.

A Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and 0 = 69.5%.
Calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as a water of crystallization. (molecular mass of the compound is 322).

30.

Define orbital ? what are the n and 1 values for 3p_x and 4d_x²_-_y² electron ?

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Answers Key:

Multiple Choice Question

15 x 1 = 15

(a)
Solids

(d)
1,2, & 3

(b)
(i) and (iv)

(d)
2NO₂ + 2OH^- \(\longrightarrow\) NO^-₂ + NO^-₃ + H₂O

(a)
+6

(a)
6.022\(\times\)10²³

10.

(b)
-1

11.

(a)
0

12.

(d)
1 moles of HCl (g)

13.

14.

15.

Generic 3

7 x 3 = 21

16.

1 x AW of N = 1 x 14 = 14amu
2 x AW of O = 2 x16 = 32 amu
Formula weight of NO₂ = 46 amu

17.

Molar mass of H₅IO₆ = (5x1 + 1x127 + 6x16)
= 228 g mol^-1
Mass of 5.14 mol of H₅IO₆ =5.14 mol x 228g mol^-1
= 1171.9 g.

18.

Molar mass of CO₂ = (1 x 12 + 2 x 16)
= 44 g mol^-1
Mass of 3 moles of CO₂ = 3 mol x 44g mol^-1
= 132 g

19.

Molar mass of NH₃= (1 x 14 + 3 x 1) = 17g mol^-1
Mass of 0.2 mol of NH₃ = 0.2 mol x 17g mol^-1
= 3.4 g

20.

No. of silver atoms = 1 million = 1 x 10⁶
Mass of one million Ag atoms = 1.79 x 10^-16g
Mass of 6.023 x 10²³atoms of silver
= \(\frac { 1.79\times { 10 }^{ -16 }g }{ 1\times { 10 }^{ 6 } } \times 6.023\times { 10 }^{ 23 }\)
= 107.8 g.
Atomic mass of silver = 6.023 x 10²³ atoms of Ag
\(\therefore\) The atomic mass of Ag = 107.8 g

21.

Compound	Given No.of moles	No.of oxygen atoms
Ethanol - C₂H₅OH	1	1\(\times\)6.022\(\times\)10²³
Formic acid - HCOOH	1	2\(\times\)6.022\(\times\)10²³
Water - H₂O	1	1\(\times\)6.022\(\times\)10²³
Formic acid

22.

Molar mass = density x Molar volume
= 1.965 x 2.24 x 10^-2
= 4.4016 x 10^-2 kg/mol
= 4.4016 x 10^-2 x 10³ g/mol
= 44.016 g/mol

Generic 2

3 x 2 = 6

23.

n	l	m	orbitals	Total no of orbitals
	0	0	1	(1- 4s +3 - 4P orbital +5 - 4d orbital +7 - 4f orbital) =16
4	1	-1 0 +1	3
	2	-2 -1 0 +1 +2	5
	3	-3 -2 -1 0 +1 +2 +3	7

24.

Energy electrons symmetry
This is due to the symmetrical distribution and exchange energy of given d- electrons. Symmetry leads to stability.
Exchange energy:
If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy. If more number of exchanges are possible, more exchange energy in released. More number of exchanges are possible only in case of half filled and fully filled configurations.
For example, in chromium the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges as shown in figure. On the other hand only six exchanges are possible for [Ar]3d⁴ 4s² configuration. Hence, exchange energy for the half filled configuration is more. This increases the stability of half filled 3d orbitals.

The exchange energy is the basis for Hund's rule, which allows maximum multiplicity, that is electron pairing is possible only when all the degenerate orbitals contain one electron each.

25.

Average atomic mass
= \(\frac { (78.99\times 23.99)+(10\times 24.99)+(11.01\times 25.98) }{ 100 } \)
= \(\frac { 2430.9 }{ 100 } \)
= 24.31 u

Generic 5

5 x 5 = 25

26.

half reaction \(\Rightarrow \) \({ S }_{ 2 }{ O }_{ 3 }^{ 2- }\longrightarrow { S }_{ 4 }{ O }_{ 6 }^{ 2- }\)
\({ I }_{ 2 }\longrightarrow { I }^{ - }\)

27.

\(\overset { +3 }{ C } _{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow \overset { +4 }{ C } { O }_{ 2 }\)
\(\overset { +6 }{ Cr } _{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { Cr }^{ 3+ }\)
(1) \(\Rightarrow \) \({ C }_{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow { 2CO }_{ 2 }+{ 2e }^{ - }\)
\({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 7H }^{ 2 }O\)

28.

Half reactions are:
\(\overset { +7 }{ M } { nO }_{ 4 }^{ - }\longrightarrow { Mn }^{ 2+ }\)
and \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }\)
(1) \(\Rightarrow \) \({ MnO }_{ 4 }^{ - }+{ 8H }^{ - }+5e^{ - }\longrightarrow { Mn }^{ 2+ }+{ 4H }_{ 2 }O\)
(2) \(\Rightarrow \) \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ 2e }^{ - }\)

ii)

iii)

iv)

29.

Element	%	Relative number of atoms	Simple Ratio
Na	14.31	\(\frac { 14.31 }{ 23 } =0.62\)	\(\frac { 0.62 }{ 0.31 } =2\)
S	9.97	\(\frac { 9.97 }{ 32 } =0.31\)	\(\frac { 0.31 }{ 0.31 } =1\)
H	6.22	\(\frac { 6.22 }{ 1 } =6.22\)	\(\frac { 6.22 }{ 0.31 } =20\)
O	69.5	\(\frac { 69.5 }{ 16 } =4.34\)	\(\frac { 4.34 }{ 0.31 } =14\)

Empirical formula = Na₂SH₂₀O₁₄
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄10H₂O.

30.

The solution to Schrodinger equation gives the permitted total energy values called eigen values and the corresponding wave function represent atomic orbitals.