Which of the following pair of compounds will have three sp^{2} -hybrid orbitals?
SO_{2}, CH_{4}
SO_{3}, C_{2}H_{4}
BF_{3}, SF_{3}
${\mathrm{I}}_{3}^{-}$, SF_{4}
B.
SO_{3}, C_{2}H_{4}
Among all the given pairs, second pair has sp^{2}- hybrid orbitals.
SO_{2} is trigonal planar. It has 3 sp^{2}- hybrid orbitals.
CH_{4} is tetrahedral. It has 4 sp^{3}- hybrid orbitals.
SO_{3} and C_{2}H_{4} are trigonalplanar and have 3 sp^{2}- hybrid orbitals.
BF_{3} is trigonal planar. It also has 3 sp^{2}- hybrid orbitals.
SF_{4} is irregular tetrahedral. It has 5 sp^{3}d hybrid orbitals.
${\mathrm{I}}_{3}^{-}$L is linear and has 5 sp^{3}d hybrid orbitals.
Work is being performed, when a weight lifter lifts a base ball off a weight rack. This is due to
magnetic attraction
gravity
electrostatic repulsion
mechanical force
B.
gravity
Due to gravity, barbell is lifed and hence work is done.
What will be the number of waves formed by a Bohr electron in one complete revolution in its second orbit?
Three
Two
One
Zero
B.
Two
Number of waves = $\frac{\mathrm{Number}\mathrm{of}\mathrm{waves}}{\mathrm{de}-\mathrm{Brogile}\mathrm{wavelength}\mathrm{of}\mathrm{electron}}$
But, $\mathrm{\lambda}=\frac{\mathrm{h}}{\mathrm{mv}}$
$\therefore $ $\mathrm{\lambda}=\frac{2\mathrm{\pi}3\times \mathrm{\mu}}{\mathrm{h}}=\frac{2\mathrm{\pi}}{\mathrm{n}}\times \mathrm{mvr}$
$\because \mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi}}=\frac{2\mathrm{\pi}}{\mathrm{h}}\times \frac{\mathrm{nh}}{2\mathrm{\pi}}$
mvr = n
For second orbit, n = 2. Therefore, number of waves = 2.
At 27°C, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. Choose the correct option from the following
Change in internal energy is positive
Heat is negative
Work done is - 965.84 cal
All are incorrect
B.
Heat is negative
Work done in isothermal reversible process is
w = -2.303 nRT ${\mathrm{log}}_{10}\left(\frac{{\mathrm{p}}_{1}}{{\mathrm{p}}_{2}}\right)$
Given, n = 1; R = 2 cal K^{-1} mol^{-1}
T = (27 + 273) K = 300 K
p_{1} = 2 atm; p_{2} = 10 atm
Therefore,
w = -2.303 $\times $ 1 $\times $ 2 $\times $ 300 ${\mathrm{log}}_{10}\left(\frac{2}{10}\right)$
w = + 965.54 cal
For isothermal change, $\u2206\mathrm{U}=0$
Now, from first law of thermodynamics,
q = $\u2206$U - w = 0 - 965.84 cal
q = -965.84 cal
Beryllium differs in properties from other elements of its own group but shows resemblance with aluminium because of
relatively bigger ionic radius and high polarising power of Be
relatively smaller ionic radius and high polarising power of Be
relatively bigger ionic radius is the only reason behind this
None of the above
B.
relatively smaller ionic radius and high polarising power of Be
Beryllium or Be differs in properties from other elements of its own group but shows resemblance with aluminium because of its smaller ionic radius and high polarising power. Polarising powers of Be^{2+} and Al^{3+} ions are almost the same thus, they show similarities in their properties.
Choose the incorrect statement about noble gas?
Boiling point increases with increasing atomic mass
Helium has least tendency to form compound
Noble gases have some value of electron affinity
Xenon has maximum number of compounds
C.
Noble gases have some value of electron affinity
Noble gases make up a group of chemical elements with similar properties. These include helium, neon, argon, krypton, xenon and radon.
Among all the given statements, statement c is incorrect. It can be corrected as valence shells of noble gases are complete, thus, they do not show tendency to gain electrons. Therefore, they have zero electron affinity.
Temperature of a gas is t K. What would be the temperature at which volume and pressure, both will reduced to half of the initial values?
$\frac{t}{2}$
$\frac{t}{4}$
$\frac{t}{3}$
$\frac{t}{8}$
B.
$\frac{t}{4}$
Given that,
p_{1} = p_{1}T_{1} = t; V_{1} = v
p_{2} = $\frac{p}{2}$; V_{2} = $\frac{v}{2}$
T = ?
By using Boyle's law
$\frac{{p}_{1}{V}_{1}}{{T}_{1}}=\frac{{p}_{2}{V}_{2}}{{T}_{2}}$
$\frac{p\times V}{t}=\frac{p\times V}{2\times 2\times {T}_{2}}$
${T}_{2}=\frac{t}{4}$
Choose the correct statement from the following
${\mathrm{NH}}_{4}^{+}$ and CH_{4} are not isoelectronic species
BF_{3} does not have dipole moment
O - Cl - O obeys octet rule
O^{-} in O_{3} is Sp^{3}-hybridised
B.
BF_{3} does not have dipole moment
Among all the given statements, statement b is correct. Other statements can be corrected as-
a. For ${\mathrm{NH}}_{4}^{+}$, number of valance electrons = 10
For CH_{4}, number of valance electrons = 10.
They both are isoelectronic species.
c. O - Cl - O does not follow octet rule.
d. O^{-} in O_{3} is sp^{2-} hybridised.
The melting point of solid substances is directly proportional to pressure acting on them. However, ice-melts at a temperature lower than its usual melting point, when the pressure increases. This is because
ice is not a true solid
the bonds break under pressure
ice is less denser than water
pressure generates heat
C.
ice is less denser than water
For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Ice melts at a temperature lower than its usual melting point because ice is less denser than water.
Aqueous solution of AlCl_{3} is acidic towards litmus while of NaCl is not. The correct reason behind this is
AlCl_{3}, furnishes OH^{-} ion in the solution
AlCl_{3}, furnishes H^{+} ion in the solution
AlCl_{3}, furnishes both H^{+} well as OH^{-} ion in the solution
AlCl_{3}, is the salt of strong base and strong acid
B.
AlCl_{3}, furnishes H^{+} ion in the solution
AlCl_{3} is the salt of weak base and strong acid which renders H^{+} ions on hydrolysis. Thus, aqueous solution of AlCl_{3} is acidic in nature.
Al^{3+ }+ 3H_{2}O $\rightleftharpoons $Al(OH)_{3} + 3H^{+}
On the other hand, NaCl is the salt of strong acid and strong base which does not undergo hydrolysis.
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