Multiple Choice QuestionsThe correct order of C-O bond length among CO, CO32-, CO2 is,
CO2< CO32-< CO
CO < CO32-< CO2
CO32- <CO2 <CO
CO < CO2 < CO32-
B.
CO < CO32-< CO2
The bond length is the average distance between the centres of nuclei of two bonded atoms. Centres of nuclei of two bonded atoms. A multiple bonds (double or triple bond) is always shorter than the corresponding single bond.
The C- atom is CO32- is sp2 hybridised as shown
The C- atom is CO2 is sp hybridised with a bond distance of carbon -oxygen is 122 pm.
The C- atom in CO is sp hybridised with C-O bond distance is 110 pm:
So the correct order is
CO < CO32-< CO2
Calculate the pOH of a solution at 250C that contains 1 x 10-10 M of hydronium ions.
7
4
9
1
B.
4
[H3O+] = [H+] = 10-10
pH + pOH = [14]
pH = - log [ H+]
pH = - log [10-10]
pH = 10
⇒ pOH + 10 = 14
⇒ pOH = 14-10 = 4
The following equilibrium constants are given:
N2 + 3H2 ⇌ 2NH3; K1
N2 +O2 ⇌ 2NO; K2
H2 + 1/2O2 ⇌ H2O' K3
The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:
K2K33 /K1
KK2K32 /K1
K22K3 /K1
K1K2 /K3
A.
K2K33 /K1
The required equation for the oxidation of NH3 by oxygen to give NO is:
With which of the following electronic configuration an atom has the lowest ionisation enthalpy?
1s2 2s22p5
1s2 2s22p3
1s2 2s22p5 3s1
1s2 2s22p6
C.
1s2 2s22p5 3s1
The electronic configuration 1s2 2s22p5 3s1 shows lowest ionisation energy because this configuration is unstable due to the presence of one electron is s- orbital. Hence, less energy is required to remove the electron.
Consider the following sets of quantum numbers:
| n | l | m | s | |
| i) | 3 | 0 | 0 | +1/2 |
| ii) | 2 | 2 | 1 | +1/2 |
| iii) | 4 | 3 | -2 | -1/2 |
| iv) | 1 | 0 | -1 | -1/2 |
| v) | 3 | 2 | 3 | +1/2 |
ii, iii and iv
i, ii, iii and iv
ii, iv and v
i and iii
C.
ii, iv and v
The value of l varies from 0 to n-i and the value of m varies from -l to +l through zero.
the value of 's' 
Which signifies the spin of the electron. The correct sets of quantum number are following:
| n | l | m | s | |
| ii) | 2 | 1 | 1 | +1/2 |
| iv | 1 | 0 | 0 | -1/2 |
| v) | 3 | 2 | 2 | +1/2 |
Which of the following oxidation states are the most characteristics for lead and tin respectively?
+4,+2
+2,+4
+4,+4
+2,+2
B.
+2,+4
The tendency to form +2 ionic state increase on moving down the group due to the inert pair effect.
Most characteristic oxidation state for lead and tin are +2,+4 respectively.
Consider the following reactions:
Enthalpy of formation of H2O (l) is:
- x2 kJ mol-1
+ x3 kJ mol-1
- x4 kJ mol-1
+ x1 kJ mol-1
A.
- x2 kJ mol-1
Enthalpy of formation: The amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent elements is known as heat of formation. SO, the correct answer is:
Given that bond energy of H-H and Cl- Cl are 430 kJ mol-1 and 240 kJ mol-1 respectively and ΔHf for HCl is -90 kJ mol-1. Bond and ΔHf for HCl is -90 kJ mol-1.Bond enthalpy of HCl is:
290 kJ mol-1
380 kJ mol-1
425 kJ mol-1
425 kJ mol-1
B.
380 kJ mol-1
An element, X has the following isotopic composition;
200X: 90%
199X : 8.0%
202X ; 2.0 %
The weighted average atomic mass of the naturally -occurring element X is closet to:
200 amu
201 amu
202 amu
199 amu
A.
200 amu
Weight of 200X = 0.90 x 200 = 180.00 amu
Weight of 199X = 8.08 x 199 = 15.92 amu
Weight 0f 202X = 0.02 x 202 = 4.04 amu
Total weight = 199.06 = 200 amu
A Weak acid, HA has a Ka of 1.00 x 10-5. If 0.100 mole of this percentage of acid dissociated at equilibrium is closest to:
99.0%
1.00%
99.9 %
0.100%
B.
1.00%
Severity: Notice
Message: Undefined variable: flex_subject_id
Filename: features/flexipad.php
Line Number: 348
Backtrace:
File: /var/www/html/public_html/application/views/features/flexipad.php
Line: 348
Function: _error_handler
File: /var/www/html/public_html/application/views/previous_year_papers/paper.php
Line: 163
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 2109
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once
