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The correct option is (c) 10 cm³.

Explanation:

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁ = (300 x 310)/300 = 310 cm³

Volume of air expelled = 310 cm³ - 300 cm³ = 10 cm³

Compounds that formally contain Pb⁴⁺ are easily reduced to Pb²⁺. The stability of the lower oxidation state is due to  Inert pair effect.

Correct option (a) 0.72,1.60

Explanation:

Atomic size of fluorine is its covalent radius, while the atomic size of neon is its van der Waals' radius.

Correct option (a) 14.6, 13.6 

Explanation:

The first ionisation potential of l{ > O because of extra stability gained by half-filled  p-orbitals of N.

N₇ = 1s², 2s², 2p¹_x, 2p_y¹ , 2p¹_z

O₈ = 1s², 2s², 2p²_x, 2p_y¹ , 2p¹_z

Correct option (a)(b)(c)

Explanation:

N₇ = 1s²,2s² 2p³

O₈ =  1s², 2s² 2p⁴

Half-filled or completely filled orbitals are more stable.

Hence. IE of oxygen is less than that of nitrogen.

Correct option (c) nitrogen

Explanation:

Generally first ionisation potential increases from left to right in a period but first ionisation potential of nitrogen is greater than that of oxyen due to presence of half-filled 2p orbitals in nitrogen.

On Mulliken scale, the average of ionisation potential and electron affinity is known as electronegativity.

False : Order of electron affinity Cl > F > Br because the size of F^-atom is lower than that of Cl^- atom, so the electron density on the surface of F is higher than that of chlorine. Thus, during the addition of an additional electron. higher repulsion takes place in fluorine than chlorine.

So, more amount of energy is consumed for the accommodation of addition electron in fluorine than chlorine. Hence, not released a large amount of energy.

This energy is higher in chlorine than fluorine ie, electron affinity of F<Cl.

Bromine shows lower electron affinity than F and Cl due to its larger size. 

Increasing order of size

Ca²⁺ < Ar < Cl^- < S^2-

Radii ∝ 1/effective nuclear charge

Answer:

(a) P’(2) will be the derivative of P(x) evaluated at x=2. So first, we take a derivative.
In this section, you should have learned product rule, so P’(x) will look like this:
P’(x)=F(x)G’(x)+F’(x)G(x)

Now we let x=2.
P’(2)=F(2)G’(2)+F’(2)G(2)

Read these values off the graph.
We see that F(2)=3. (When its x-value is 2, the curve F has a y-value of 3)
We see that G(2)=2.
F’(2) will be the slope of the curve F at 2. A tangent line at F(2) would be horizontal, so F’(2)=0.
G(2) has a slope of 1/2, so G’(2)=1/2

Now we substitute these values into P’(2) to solve:

P’(2)=(3)(1/2)+(0)(2)=(3/2)+(0)=3/2

(b) We’re playing by the same rules as above–differentiate the equation, read the values off the graph, substitute the values into the function, and solve.

Differentiate (Quotient Rule):
Q’(x)=[G(x)F’(x)-F(x)G’(x)]/[G(x)²]
Read the values off the graph:
G(7)=1
G’(7)=-2/3
F(7)=5
F’(7)=1/4
Substitute:
Q’(7)=[1*(1/4)-5*(-2/3)]/[1²]
Q’(7)=[(1/4)+(10/3)]
Q’(7)=(43/12)=3.5833.

I hope it would help you to get the answer!!!

We have, for dodecagon sin(2π/40) = (a/2)/r     

so, r = (a/2)/ sin(π/20)

and for hexagon sin(2π/12) = (x/2)/r       

so, x/2 = r sin(π/6)  =  (a/2) sin(π/6)/ sin(π/20)  

Now solve for x, you'll get the answer.

|a|<1

Option (iii) is the correct answer.

You can calculate n factor in disproportionation reaction by the formula

 1/N=1/n1 +1/n2

where n₁=n factor for oxidation half  and n₂=n factor for reduction half

As example:
I₂ → IO₃^- + I⁻
Now we calculate n₁ -
The oxidation reaction is I₂ → 2IO₃⁻ + 10e⁻
n₁= 10
Now we calculate n₂ –
The reduction reaction is - 2e⁻ + I₂ → 2I^- 
n₂ = 2
Hence , n = 5/3

Answer: A,B

Amides can be hydrolysed in both acidic medium. During hydrolysis of amides, alkylnitrogen bond is not dissociated, hence configuration at a-carbon of amine is retained.

Ca²⁺ has a smaller ionic radius than K⁺ because it has Higher effective nuclear charge

The energy released when an electron is added to a neutral gaseous atom is called Electron affinity of the atom.

Its look like your question is not complete please post your complete question..............

Correct option (d) For transition elements the d-subshells are tilled with electrons monotonically with increase in atomic number.

Explanation:

Among transition elements. electrons are not filled in d-subshell monotonically with increasing atomic number.

Correct option (b) [Ne] 3s² 3p³

Explanation:

Ionisation energy increases with increasing atomic number in a period. while decreases on moving down the group.

Correct option (d) N^a+ 

Explanation:

For isoelectronic ions

ionic size ∝ 1/atomic number

Correct option (a) Na <Mg>AI< Si

Explanation:

First ionisation potential increases from left to right across a period but Mg has extra stability than Al, due to full-filled 3s -orbitals.

Na₁₁  = 1s²,2s²,2p⁶,3s¹

Mg₁₂  = 1s²,2s²,2p⁶,3s²

AI₁₃  =  1s²,2s²,2p⁶,3s², 3p¹

Si₁₄  = 1s²,2s²,2p⁶,3s², 3p²

Thus, correct order of first ionisation potential is :

Na<Mg>AL<Si

The possible sets are {±2 ,± 3} and {±4, ±1};

Therefore, number of elements in the required set is 8.