NEET-UG 2011

${\left({\mu }_0{\epsilon }_0\right)}^{-1/2}$ = $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ = Speed of light in vacuum = c

So the dimension of ${\left({\mu }_0{\epsilon }_0\right)}^{-1/2}$ = [c] =[LT^-1]

We know, v² = u² + 2gh

Here u = 0

$\therefore$ v = $\sqrt{2gh}$ = $\sqrt{2\times 10\times 20}$ = 20 m/s

Velocity towards east direction, $\overrightarrow{v_1}=30\hat{i}$ m/s

Velocity towards north direction, $\overrightarrow{v_2}=40\hat{j}$ m/s

Change in velocity, $\Delta \overrightarrow{v}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1$ = $\left(40\hat{j}-30\hat{i}\right)$

$\therefore \left|\Delta \overrightarrow{v}\right|=\left|40\hat{j}-30\hat{i}\right|=50m/s$

Average acceleration, ${\overrightarrow{a}}_{av}$ = $\frac{changeinvelocity}{Timeinterval}$

${\overrightarrow{a}}_{av}=\frac{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}{\Delta t}=\frac{\Delta \overrightarrow{v}}{\Delta t}$

$\left|{\overrightarrow{a}}_{av}\right|=\frac{|\Delta \overrightarrow{v}|}{\Delta t}=\frac{50m/s}{10s}=5m/s^2$

For maximum range, the angle of projection, $\theta$ = 45°.

$\therefore R=\frac{u^2\sin 2\theta }{g}$

$=\frac{{\left(20\right)}^2\sin \left(2\times {45}^{\circ }\right)}{10}$

$=\frac{400\times 1}{10}=40m$

Centripetal acceleration $a_c={\omega }^{2}r$

$={\left(\frac{2\pi }{T}\right)}^{2}r$

$={\left(\frac{2\pi }{0.2\pi }\right)}^2\times 5\times 10^{-2}=5$ m/s²

Impulse experienced by the body = change in momentum
= MV – (–MV)
= 2MV.

Here, Mass of a person, m = 60 kg Mass of lift, M = 940 kg,

a = 1 m/s², g = 10 m/s²

Let T be the tension in the supporting cable.

$\therefore$ T – (M + m)g = (M + m)a

$\Rightarrow$ T = (M + m)(a + g)= (940 + 60)(1 + 10) = 11000 N

Power, P = $\overrightarrow{F}.\overrightarrow{v}=Fv\cos \theta$
Just before hitting the earth $\theta$ = 0°. Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.

Work done = Area under (F-d) graph

= Area of rectangle ABCD + Area of triangle DCE

= $2\times \left(7-3\right)+\frac{1}{2}\times 2\times \left(12-7\right)=8+5=13J$

When angular acceleration (a) is zero then torque on the wheel becomes zero.

$\theta \left(t\right)=2t^3-6t^2$

$\Rightarrow \frac{d\theta }{dt}=6t^2-12t$

$\Rightarrow \alpha =\frac{d^2\theta }{d{t}^2}=12t-12=0$

$\therefore$ t = 1 sec.

By the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia about a parallel axis through the centre of mass and Ma² , where M is the mass of the body and a is the separation between the two axes.

Therefore, the moment of inertia of the rod about an axis passing through one of its ends and perpendicular to its length is

I = I₀ + M (L/2)²

= I₀ + ML²/4

Angular momentum is conserved

$\therefore$ L₁ = L₂

$\Rightarrow$ mr₁v₁ = mr₂v₂

$\Rightarrow$ r₁v₁ = r₂v₂

$\Rightarrow \frac{v_1}{v_2}=\frac{r_2}{r_1}$

Change in entropy is given by

$dS=\frac{dQ}{T}\Rightarrow \Delta S=\frac{\Delta Q}{T}=\frac{m{L}_f}{273}$

$\Delta S=\frac{1000\times 80}{273}=293cal/K$

Note : In the question paper unit of latent heat of ice is given to be cal/°C. It is wrong. The unit of latent heat of ice is cal/g.

If a process is expansion then work done is positive so answer will be (a).

But in question work done by gas is given –150J so that according to it answer will be (d).

y = sin$\omega$t – cos$\omega$t

$=\sqrt{2}\left[\frac{1}{\sqrt{2}}\sin \omega t-\frac{1}{\sqrt{2}}\cos \omega t\right]$

$=\sqrt{2}\sin \left(\omega t-\frac{\pi }{4}\right)$

It represents a SHM with time period, $T=\frac{2\pi }{\omega }$

$y={\sin }^3\omega t=\frac{1}{4}\left[3\sin \omega t-\sin 3\omega t\right]$

It represents a periodic motion with time period

$T=\frac{2\pi }{\omega }$ but now SHM.

$y=5\cos \left(\frac{3\pi }{4}-3\omega t\right)$

$=5\cos \left(3\omega t-\frac{3\pi }{4}\right)$   $\left[\because \cos \left(-\theta \right)=\cos \theta \right]$

It represents a SHM with time period, $T=\frac{2\pi }{3\omega }$

$y=1+\omega t+{\omega }^2{t}^2$

It represents a non-periodic motion. Also it is not physically acceptable as y $\to$ $\infty$ as t $\to$ $\infty$.

Here, y₁ = a sin ($\omega$t + kx + 0.57)
and y₂ = a cos ($\omega$t + kx)

$=a\sin \left[\frac{\pi }{2}+\left(\omega t+kx\right)\right]$

Phase difference, $\Delta \varphi ={\varphi }_2-{\varphi }_1$

$=\frac{\pi }{2}-0.57$

$=\frac{3.14}{2}-0.57$

= 1.57 – 0.57 = 1 radian

We have, v = n$\lambda$

$\Rightarrow v\propto \lambda$ (as n remains constant)

Thus, as v increases 10 times, $\lambda$ also increases 10 times.

Distance of point A from the two +q charges = L.

Distance of point A from the two –q charges
$=\sqrt{L^2+{\left(2L\right)}^2}=\sqrt{5}L$



$\therefore$ $V_A=\left(\frac{Kq}{L}\times 2\right)-\left(\frac{Kq}{\sqrt{5}L}\times 2\right)$

$=\frac{2Kq}{L}\left[1-\frac{1}{\sqrt{5}}\right]$

$=\frac{1}{4\pi {\epsilon }_0}.\frac{2q}{L}\left(1-\frac{1}{\sqrt{5}}\right)$

According to Gauss’s law

${\varphi }_E=\frac{Q_{enclosed}}{{\epsilon }_0}$

If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same. This is because electric flux depends only on the charge enclosed by the surface.

Let $\epsilon$ be the emf and r be internal resistance of the battery.

In the first case,


$2=\frac{\epsilon }{2+r}$

In the second case,


$0.5=\frac{\epsilon }{9+r}$

Divide (i) by (ii), we get

$\frac{2}{0.5}=\frac{9+r}{2+r}\Rightarrow 4+2r=4.5+0.5r$

$1.5r=0.5\Rightarrow r=\frac{0.5}{1.5}=\frac{1}{3}\Omega$

We have,

$P=\frac{V^2}{R}$

$\Rightarrow 36=\frac{V^2}{9}$

$\Rightarrow V=18$

Current passing through the 9$\Omega$ resistor is

$i_1=\frac{V}{R}=\frac{18}{9}=2A$

The 9$\Omega$ and 6$\Omega$ resistors are in parallel, therefore

$i_1=\frac{6}{9+6}\times i$

where i is the current delivered by the battery.

$\therefore i=\frac{2\times 15}{6}=5A$

Thus, potential difference across 2$\Omega$ resistor is

V = iR = 5 × 2 = 10V

Capacitance of a parallel plate capacitor is

$C=\frac{{\epsilon }_{0}A}{d}$   ...(i)

Potential difference between the plates is

V = Ed    ...(ii)

The energy stored in the capacitor is

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{{\epsilon }_{0}A}{d}\right){\left(Ed\right)}^2$   (Using (i) and (ii))

$=\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

Here, ${\overrightarrow{F}}_{BC}=\overrightarrow{F}$

$\because {\overrightarrow{F}}_{AB}=0$

The net magnetic force on a current carrying closed loop in a uniform magnetic field is zero.

$\therefore {\overrightarrow{F}}_{AB}+{\overrightarrow{F}}_{BC}+{\overrightarrow{F}}_{AC}=0$

$\Rightarrow {\overrightarrow{F}}_{AC}=-{\overrightarrow{F}}_{BC}$    ($\because$ ${\overrightarrow{F}}_{AB}=0$)

$\overrightarrow{v}$ and $\overrightarrow{B}$ are in same direction so that magnetic force on electron becomes zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.

Diamagnetic will be feebly repelled.

Paramagnetic will be feebly attracted.

Ferromagnetic will be strongly attracted.

Therefore, A is of diamagnetic material. B is of paramagnetic material. C is of ferromagnetic material. D is of non-magnetic material.

e = -L$\frac{di}{dt}$

During 0 to $\frac{T}{4}$, $\frac{di}{dt}$ = constant

$\therefore$ e = -ve

During $\frac{T}{4}$ to $\frac{T}{2}$, $\frac{di}{dt}$ = 0

$\therefore$ e = 0

During $\frac{T}{2}$ to $\frac{3T}{4}$, $\frac{di}{dt}$ = 0

$\therefore$ e = +ve

Thus graph given in option (a) represents the variation of induced emf with time.

tan $\varphi$ = $\frac{X_L}{R}$ = $\frac{3}{3}$ = 1

$\Rightarrow$ $\varphi$ = 45^o or $\frac{\pi }{4}$

e₀ = $200\sqrt{2}$ V, $\omega$ = 100 rad s^-1

The capacitive reactance is

X_C = $\frac{1}{\omega C}=\frac{1}{100\times 1\times {10}^{-6}}$

The r.m.s. value of the current in the circuit is

i_r.m.s = $\frac{V_{rms}}{X_C}=\frac{\frac{e_0}{\sqrt{2}}}{\frac{1}{\omega C}}$

= $\frac{\frac{200\sqrt{2}}{\sqrt{2}}}{\frac{1}{100\times {10}^{-6}}}$

= 200 × 100 × 10^–6 A = 2 × 10^–2 A = 20 mA

The e.m. wave is propagating along +z direction.

Electric field and magnetic field are mutually perpendicular and also perpendicular to the direction of propagation.

$\therefore$ $\left[\overrightarrow{E}=E_0\hat{i},\overrightarrow{B}=B_0\hat{j}\right]$

From the electromagnetic wave spectrum, the correct decreasing order of wavelength is :

microwave, infrared, ultraviolet, gamma rays.

Difference between apparent and real depth of a pond is due to refraction. Other three are due to total internal reflection.

Focal length of the lens

$\frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{20}-\frac{1}{-20}\right)$

$\Rightarrow$ f = 20 cm

From lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow$ $\frac{1}{v}-\frac{1}{-30}=\frac{1}{20}$

$\Rightarrow$ v = 60 cm

$\frac{I}{O}=m=\frac{v}{u}$ = $\frac{60}{-30}$ = -2

$\therefore$ I = -2 $\times$ (O) = -2 $\times$ 2 = -4 cm

So image will be real inverted and of size 4 cm.

Extremely high temperature needed for fusion make kinetic energy large enough to overcome coulomb repulsion between nuclei.

$\alpha$-particle is ₂He⁴

In $\beta$ emission, the neutron gets converted to proton and electron

_nX^m $$\frac{\text{buildrel}1\alpha }{⟶}$$ _{n - 2}Y^{m - 4} $$\frac{\text{buildrel}2\beta }{⟶}$$ _nZ^{m - 4}

Momentum Mu = $\frac{E}{c}=\frac{hv}{c}$

Recoil energy :

$\frac{1}{2}M{u}^2$ = $\frac{1}{2}\frac{M^2{u}^2}{M}$

= $\frac{1}{2M}{\left(\frac{hv}{c}\right)}^2$

= $\frac{h^2{\nu }^2}{2Mc}$

From Einstein relation, E = mc²

Rearranging, m = $\frac{E}{c^2}$

We see that mass decay per second :

$\frac{dm}{dt}$ = $\frac{1}{c^2}$ × $\frac{dE}{dt}$ = $\frac{1000\times {10}^3}{{\left(3\times {10}^8\right)}^2}$

Now mass decay of U²³⁵ per hour,

= $\frac{dm}{dt}$ $\times$ 60 $\times$ 60

= $\frac{1000\times {10}^3}{{\left(3\times {10}^8\right)}^2}$ $\times$ 3600

= 4 × 10^–8 kg = 40 microgram

Initial number of atoms of X is N₀

and Initial number of atoms of Y is 0

Number of atoms after time t, for X is N and for Y is N₀ - N

According to the question,

$\frac{N}{N_0-N}$ = $\frac{1}{15}$

$\Rightarrow$ $\frac{N}{N_0}=\frac{1}{16}$

As $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$ where n is the no. of half lives

$\therefore$ ${\left(\frac{1}{2}\right)}^n=\frac{1}{16}$

$\Rightarrow$ n = 4

t = nT_1/2 = 4 × 50 = 200 years

Hence, the age of rock is 200 years.

The wavelength of the first line of lyman series for hydrogen atom is

$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

The wavelength of the second line of Balmer series for hydrogen like ion is

$\frac{1}{{\lambda }^{\prime }}=R{Z}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

According to question

$\lambda$ = $\lambda$'

$R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ = $R{Z}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

$\Rightarrow$ $\frac{3}{4}$ = $\frac{3Z^2}{16}$

$\Rightarrow$ Z = 2

The de Broglie wavelength $\lambda$ associated with the electrons is

$\lambda =\frac{1.227}{\sqrt{V}}$

where V is the accelerating potential in volts.

$\therefore$ $\lambda$ $\propto$ $\frac{1}{\sqrt{V}}$

$\Rightarrow$ $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{V_2}{V_1}}$

= $\sqrt{\frac{100\times {10}^3}{25\times {10}^3}}$ = 2

$\Rightarrow$ ${\lambda }_2=\frac{{\lambda }_1}{2}$

Maximum K.E. = Stopping Potential

So stopping potential will be 0.5 V

Davisson and Germer experiment describes that velocity of electron emitted from electron gun increases using potential difference between anode and filament.

According to Einsten’s photoelectric effect, the K.E. of the radiated electrons

K.E_max = E – W

$\frac{1}{2}m{v}_1^2$ = (1 - 0.5) eV = 0.5 eV

$\frac{1}{2}m{v}_2^2$ = (2.5 - 0.5) eV = 2 eV

$\frac{v_1}{v_2}=\sqrt{\frac{0.5}{2}}$ = 1 : 2

According to Einstein’s photoelectric equation

K_max = h$\nu$ – h$\nu$₀

Since K_max is +ve, the photoelectric emission occurs only if

h$\nu$ $>$ h$\nu$₀ or $\nu$ $>$ $\nu$₀

The photoelectic emission occurs only when the incident light has more than a certain minimum frequency. This minimum frequency is called threshold frequency.

In forward biasing, the positive terminal of the battery is connected to p-side and the negative terminal to n-side of p-n junction. The forward bias voltage opposes the potential barrier. Due to it, the depletion region becomes thin.

If a small amount of antimony is added to germanium crystal, crystal becomes n-type semiconductor. Hence, there will be more free electrons than holes.

AND, NAND and NOT gates are (iii), (iv) and (ii) respectively.

Current gain, $\beta$ = $\frac{\Delta I_C}{\Delta I_B}$

= $\frac{\left(20-10\right)\times {10}^{-3}}{\left(300-100\right)\times {10}^{-6}}$ = 50

E₁ = $\frac{hc}{{\lambda }_1}$ and E₂ = $\frac{hc}{{\lambda }_2}$

$\frac{E_1}{E_2}$ = $\frac{hc}{{\lambda }_1}\times \frac{{\lambda }_2}{hc}$ = $\frac{{\lambda }_2}{{\lambda }_1}$

$\Rightarrow$ $\frac{25}{50}$ = $\frac{{\lambda }_2}{{\lambda }_1}$

$\Rightarrow$ $\frac{1}{2}$ = $\frac{{\lambda }_2}{{\lambda }_1}$ $\Rightarrow$ ${\lambda }_1$ = 2${\lambda }_2$

The total number of atomic orbitals in any energy level is n².

$\therefore$ 4² = 16

Given V₁ = 55 mL, V₂ = ?

P₁ = 715 – 15 = 700 mm, P₂ = 760 mm

T₁ = 300 K, T₂ = 273 K

General gas equation,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\Rightarrow$ $\frac{700\times 55}{300}=\frac{760\times V_2}{273}$

$\Rightarrow$ V₂ = 46.098 mL

Now, 22400 mL of nitrogen = 1 mole

$\therefore$ 46.098 mL of nitrogen = $\frac{1\times 46.098}{22400}$ mol

Weight of nitrogen = $\frac{1\times 46.098}{22400}$ $\times$ 28 = 0.057 g

Percent composition of nitrogen in 0.35 g of compound

= $\frac{0.0576}{0.35}\times 100$ = 16.45 %

We know that $\frac{r_A}{r_B}=\frac{v/t_A}{v/t_B}=\sqrt{\frac{M_B}{M_A}}$

$\Rightarrow \frac{t_A}{t_B}=\sqrt{\frac{M_B}{M_A}}\Rightarrow \frac{10}{20}=\sqrt{\frac{M_B}{49}}$

$\Rightarrow {\left(\frac{10}{20}\right)}^2=\frac{M_B}{49}\Rightarrow \frac{100}{400}=\frac{M_B}{49}$

$\Rightarrow M_B=\frac{49\times 100}{400}=12.25$ u

Average velocity (v_AV) = $\sqrt{\frac{8RT}{\pi M}}$

$\Rightarrow$ v_AV $\propto$ $\sqrt{T}$

$\therefore$ $\frac{{\left(v_{AV}\right)}_2}{{\left(v_{AV}\right)}_1}=\sqrt{\frac{2T}{T}}$ = 1.4

Number of moles $\Rightarrow$ n_CO = n_N2

Volume of container is same. So, V_CO = V_N2 .

Also, temperature is same for both the gases thus, T_CO = T_N2

According to ideal gas equation, PV = nRT

Now, V, n, R and T for both gases are same. So, P_CO = P_N2

Now, total pressure is 1 atm and according to Dalton’s law of partial pressure,

P_CO + P_N2 = 1 atm

$\Rightarrow$ 2P_N2 = 1 atm {$\because$ P_CO = P_N2 }

$\Rightarrow$ P_N2 = 0.5 atm

X_2(g) + 4Y_2(g) $\rightleftharpoons$ 2XY_4(g)

$\Delta$n_g = -ve and $\Delta$H = -ve

As $\Delta$H < 0 i.e., the given reaction is exothermic. According to Le-Chatelier principle, for exothermic reaction, forward reaction is favoured when temperature becomes low. Also, there are 5 gaseous moles on reactant side and 2 gaseous moles on products side. So, forward reaction is favoured when pressure of the reaction mixture becomes high. The reason is that at high pressure reaction tends to more in direction where there is lessee number of gaseous moles.

N_2(g) + $\frac{1}{2}$O_2(g) $\rightleftharpoons$ NO_(g),     ${\left(K_1\right)}^{\frac{1}{2}}$

NO_(g) + $\frac{1}{2}$O_2(g) $\rightleftharpoons$ NO_2(g),     ${\left(K_2\right)}^{\frac{1}{2}}$
---------------------------------------------------
$\frac{1}{2}$N_2(g) + O_2(g) $\rightleftharpoons$ NO_2(g)    ${\left(K_1{K}_2\right)}^{\frac{1}{2}}$

So, for the reverse reaction which is the desired one the value of K will be reciprocal of this value. i.e.

K = ${\left[\frac{1}{K_1{K}_2}\right]}^{1/2}$

pOH = pK_b + log $\frac{\left[Salt\right]}{\left[Base\right]}$

= – log K_b + log $\frac{\left[Salt\right]}{\left[Base\right]}$

= – log 1.8× 10^–5 + log $\frac{0.20}{0.30}$

= 5 – 0.25 + (–0.176) = 4.57

Now, pH = 14 – pOH = 14 – 4.57 = 9.43

The species which have a lone pair of electrons to donate or a negative charge on it can act as a Lewis base. Here, N atom in NH₃ and O atom in H2O have lone pair of electrons available for donation. In OH^– the negative charge on it results in to behave it as a Lewis base. But BF₃ is an electron deficient species thus, it is least likely to behave as Lewis base.

According to depression in freezing point,

$\Delta$T_f = i $\times$ K_f $\times$ m

= i $\times$ K_f $\times$ $\frac{w_B\times 1000}{m_B\times w_A}$

Given : $\Delta$T_f = 3.82, K_f = 1.86,

w_B = 5, m_B = 142, w_A = 45

i = $\frac{\Delta T\times m_B\times w_A}{K_f\times w_B\times 1000}$

= $\frac{3.82\times 142\times 45}{1.86\times 5\times 1000}$

= 2.63

From the value of van’t Hoff factor i it is possible to determine the degree of dissociation or association. In case of dissociation, i is greater than 1 and in case of association i is less than 1.

H₂O(l) $\frac{\text{buildrel}300K}{⟶}$ H₂O(g)

∆H = 30 kJ mol^–1

$\Delta S=\frac{\Delta H}{T}$ = $\frac{30\times {10}^3}{300}$

= 100 J mol^–1 K^–1

4H(g) → 2H₂(g),    ∆H = – 869.6 kJ

Reverse the above equation

2H₂(g) → 4H(g),    ∆H = + 869.6 kJ

Divide the above equation by 2,

H₂(g) → 2H(g), $\Delta H=\frac{869.6}{2}$ kJ = 434.8 kJ

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus, $\Delta$T = 0.

Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.

E°_{Sn⁴⁺/Sn²⁺} = 0.15 V

E°_Cr³⁺/Cr = –0.74 V

E°_cell = E°_cathode – E°_anode

= 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V

As the electrode potential drops, reducing power increases.

So, Z (–3.0 V) > X (–1.2 V) > Y (+ 0.5 V)

Cu²⁺_(aq) + e^$$-$$ $\to$ Cu⁺_(aq) ; E₁^o = 0.15 V

Cu⁺_(aq) + e^$$-$$ $\to$ Cu_(s) ; E₂^o = 0.50 V

Cu²⁺ + 2e^– $\to$ Cu ; E^o = ?

$\Delta$G^o = $\Delta$G₁° + $\Delta$G₂°

$\Rightarrow$ – nFE° = – n₁FE₁° – n₂FE₂°

$\Rightarrow$ E^o = $\frac{1\times 0.15+1\times 0.50}{2}$ = 0.325 V

Order of a reaction is not always whole number. It can be zero, or fractional also.

$\frac{x}{m}=p\times T$ is incorrect.

Increasing order of bond length is

$\underset{107pm}{C-H}<\underset{134pm}{C=C}<\underset{141pm}{C-O}<\underset{134pm}{C-C}$

$O_2^{+}$ and $O_2$ are paramagnetic in nature as they contain one and two unpaired electrons respectively.

Electronic configuration
O₂ : $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_y^2{\pi }^{\ast }2p_y^1{\pi }^{\ast }2p_z^1$

Bond order = $\frac{1}{2}\left(8-4\right)$ = 2

O₂⁺ : Bond order = $\frac{1}{2}\left(8-3\right)$ = 2.5

O₂^- : Bond order = $\frac{1}{2}\left(8-5\right)$ = 1.5

O₂^2- : Bond order = $\frac{1}{2}\left(8-6\right)$ = 1

As bond order increases, bond length decreases.

Van Arkel method is used for purification of Zr and Ti.

Pig iron contains carbon (nearly 4%) as the impurity to the maximum extent. Other impurities are S, P, Si, Mn in trace amount.

As the covalent character in compound increases and ionic character decreases, melting point of the compound decreases. So, CaI₂ has the highest covalent character and lowest melting point.

Active ingredient in bleaching powder for bleaching action is Ca(OCl)₂, i.e., calcium hypochlorite.

In case of Lassaigne’s test of halogens, it is necessary to remove NaCN and Na₂S from the sodium extract, if nitrogen and sulphur are present. This is done by boiling the sodium extract with conc. HNO₃.

NaCN + HNO₃ $\to$ NaNO₃ + HCN $\uparrow$

Na₂S + 2HNO₃ $\to$ 2NaNO₃ + H₂S $\uparrow$

Pyrosilicate contains two units of SiO₄^4– joined along a corner containing oxygen atom.

K₂Cr₂O₇ + 3Na₂SO₃ + 4H₂SO₄ $\to$

3Na₂SO₄ + K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O

The order can be explained using the idea of spin correlation. Spin correlation refers to lowering of energy for like (parallel) spins. Spin correlation leading to decrease in repulsion for electrons of like spins than for electrons of different spins is called exchange energy.

Spin correlation and its exchange energy gives an electronic configuration a special stability which is greatest for half-filled electronic configurations.

Mn²⁺ (d⁵ ) gets stabilisation due to half-filled configuration.

In Fe²⁺ (d⁶) the placing of one extra electron in a subshell destabilises. Placing of 2 electrons in Co²⁺ (d⁷) destabilises it more. Cr²⁺ (d⁴) has one vacant subshell. Fe²⁺ gets more stabilisation compared to Cr²⁺ through exchange energy. So the order is as follows:

Mn > Fe > Cr > Co

Complexes of the type M_ABCD may exist in three isomeric forms.



Similarly, [Pt(Py)(NH₃)BrCl] may exist in three isomeric form in which

M = Pt, A = Py, B = NH₃, C = Br, D = Cl.

Ti³⁺ $\to$ [Ar] 3d¹4s⁰ (1 unpaired electron)

Cr³⁺ $\to$ [Ar] 3d³ 4s⁰ (3 unpaired electrons)

Co³⁺ $\to$ [Ar] 3d⁶ 4s⁰ (0 unpaired electrons because of pairing)

Zn²⁺ $\to$ [Ar] 3d¹⁰ (no unpaired electrons)

[Cr(NH₃)₆]³⁺ exhibits highest paramagnetic behaviour as it contains 3 unpaired electrons.

As + ve charge on the central metal atom increases, the less readily the metal can donate electron density into the $\pi$* orbitals of CO ligand to weaken the C – O bond. Hence, the C – O bond would be strongest in Mn(CO)₆⁺.

For [Ni(CN)₄]^2–

Ni⁺² = 3d⁸4s⁰

CN– is a strong field ligand thus, it pair up the electrons.

$\therefore$ [Ni(CN)₄]^–2 is diamagnetic in nature.