Chemistry

Given are the oxide of alkali and alkaline earth metals which are ionic in nature.

Simple oxide are Li₂O, CaO, MgO and K₂O.

Peroxide is Na₂O₂ and superoxide is KO₂.

All simple oxides are diamagnetic as it has no unpaired electron.

Zn (30) = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}$  

$Z{n}^{+}=\left[Ar\right]4s^{1}3d^{10}$      

Outermost electron is 4s electron,

$n=4,\mathcal{l}=0,m=0,s=\pm 1/2$         

log k = 20.35 - $\frac{\left(2.47*10^3\right)}{T}$  

Comparing with,

$logk=logA-\frac{Ea}{2.303RT}$

$\frac{Ea}{2.303R}=2.47*10^3$

= 47.29 kJ/mole

Ans. = 47

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

$m=\frac{w*1000}{M*W_{solvent}}$

For acetone solution,

$0.17=1*1.7*\frac{1.22*1000}{M*100}$

For Benzene solution,

$2Acid\to {\left(Acid\right)}_2\therefore i=1/2$

$\Delta T_b=i*K_b*m$

$=\frac{1}{2}*2.6*\frac{1.22*1000}{122*100}°C$

= $0.13°C=13*10^{-2}°C$ ${}^{}$

$\therefore x*10^{-2}=13*10^{-2}$

$\therefore x=13$

Effective number of atom in C.C.P

$\therefore =\frac{1}{8}*8+\frac{1}{2}*6$     

= 4

Number of octahedral void = 4

Number of cations = 4

Number of anion = 4

Formula of compound = A₄B₄

Empirical formula = AB

Ans. x = 1

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H₂O

30 * 1 mmol      50 * 1 mmol     

0 mmol               20 mmol

${\left[H^{+}\right]}_{mix}=\frac{20}{50+30}M=\frac{20}{80}M=\frac{1}{4}M=0.25M$

$x*10^{-4}=6021*10^{-4}$    

$\therefore$ x = 6021

Ans. = 6021

$O_2^{-2}=8*2+2=18e^{-}$

${\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2={\pi }_{2py}^2{\pi }_{2px}^{*2}={\pi }_{2py}^{*2}$      

Number of unpaired e^- = 0

Ans. = 0

$\left(\frac{x}{m}\right)=K{P}^{1/n}$           

$\frac{10}{1}=K{\left(100\right)}^{1/n}................\left(1\right)$     

$\frac{15}{1}=K{\left(200\right)}^{1/n}................\left(2\right)$

$\frac{V}{1}=K{\left(300\right)}^{1/n}...............\left(3\right)$

Dividing (2) by (1)

$\frac{15}{10}={\left(\frac{2}{1}\right)}^{1/n}$ = (2)^1/n

$\frac{V}{10}=10^{0.279}$

$V=10*10^{0.279}=10^{1.279}=10^x$

$\therefore x=1.279=127.91*10^{-2}\approx 128*10^{-2}$

Ans. = 128

$C_2{H}_{7}N+\left(2x+\frac{y}{2}\right)CuO\to xC{O}_2+\frac{y}{2}{H}_{2}O+\frac{Z}{2}{N}_2+\left(2x+\frac{y}{2}\right)Cu$

$C_2{H}_7{N}_1+\left(2*2+\frac{7}{2}\right)CuO\to 2C{O}_2+\frac{7}{2}{H}_{2}O+\frac{1}{2}{N}_2+\left(2*2+\frac{7}{2}\right)Cu$              

$\therefore y=7$              

Ans. = 7

Students should prepare complete syllabus when they have time to prepare. However, you can use the list of highweightage chapters in last minute revision for scoring well.

• The p-Block Elements: This chapter holds a high weightage of 8–10 marks. (in the latest syllabus this is deleted)

• Aldehydes, Ketones, and Carboxylic Acids: This chapter contributes around 8–10 marks.

• Biomolecules: This chapter accounts for around 8 marks.

• Chemical Kinetics: This chapter holds a high weightage of 5-6 marks. 

• The d- and f-Block Elements: This chapter contributes around 5-6 marks.

• Amines: This chapter contributes around 5-6 marks.