This option is correct because radius of n^th orbit is given by , r _n= ( 0.529 n² ) / Z . Radius of first bohr's orbit is , r  =  0.529 . Hence, radius of n^th orbit will be = r n² . Hence, this option is correct  .

This option is correct because it is clearly seen that n = 5. In this case l value of ' p ' orbital is 1 . Hence, this option is correct .

This option is correct because  L value of ' p ' orbital is 1 . Hence, this option is correct .

This option is correct because;  , λ = h / mv . and from conservation of angular momentum we have = mvr = nh / 2π . => 2πr = nλ . in this case value of n is 4 . hence, this option is correct.

This option is correct because uncertainity principal states that the minimun uncertainity will be  h / 4 π . Hence, if uncertainity in one quantity is 0 then uncertainty of other quantity must be infinite . Hence, this option is correct .

this option is correct because , l value of ' d ' orbital is 2 hence this option is correct .

This option is correct because ,

energy is given by , En = -13.6 / n² eV / atom .

En -En-1 = -13.6 [ (1 / n² ) - (1 / (n-1)² ) ] .

This diffrence is decreases with the increase in value of ' n ' .

Hence, this option is correct .

this option is correct because , ' p ' orbital shows this shape whose l value is  = 1  . hence this option is correct .

this option is incorrect because , according to n+l rule , highest energy is possesed by that set which has highest value of n + l . so , in this case , n = 3 , l = 2 . hence n+l = 5 . which is not the highest among the options . hence this option is incorrect .

this option is correct because , each value of m implies 1 orbital and each orbital can only accommodate 2 electrons . hence this option is correct .

this option is correct because , value of ' m '  are from  - l to  + l including 0 . and l value of ' d ' orbital is 2 . hence this option is correct .

this option is correct because , lithium has electronic configuration - 1s² 2s¹ . hence according  to the question n = 2 means principle quantum number is 2  . l =0 means ' s ' orbital . s = + 1 / 2 means 1 electrons . all these characters are exhibited by lithium . therefore this option is correct .

this option is correct because , for every value of  ' m ' only one orbital exist  . hence this option is correct  .

this option is correct because , p subshell has 3 orbital each can accommodate 2 electrons hence ' p '  orbital can accommodate 6 electrons . in this case p has 7 electrons . hence this statement is wrong and therefore  this option is correct .

this option is correct because , according to pauli exclusion principal - no two electrons in an atom can have same sets of 4 quantum number . this shows that no 2 electrons in an atom are alike or orbital can be accommodate a maximum of two electrons that too with opposite spin . hence this option is correct .

this option is correct because , l value 2 is shown by ' d ' orbital . this configuration represent an electron in 4d   orbital . hence this option is correct .

This option is correct as we know that primary standards are neither hygroscopic nor deliquescent. This can be explained as follows: these are the reagents whose weight is true representative of the number of moles of the substance contained so if they would have been hygroscopic or deliquescent then they will undergo some change in their properties and will not represent the true no. of moles of the substance contained and hence these substances cannot be deliquescent of hygroscopic. Hence, this option is correct.

this option is correct because , E₁ / E₂ = Z₁² / Z₂² x n₂² / n₁²  . since n values is 1 in this question . and  Z₁ = 1 ,  Z₂ = 4 . E₁ / E₂ = 1 / 16 . hence this option is correct  . 

This option is correct and can be explained as follows: In titration, between oxalic acid and caustic soda of sodium hydroxide we find that oxalic acid is weak acid while caustic soda is a strong base hence equivalence point will lie at pH above 8 or basic. We know that phenolphthalein gives a transition from colourless to fuchsia colour at pH above 8.2. Hence, we can use phenolphthalein indicator for this reaction.

This option is correct and can be explained a follows: In the titration of sodium carbonate with hydrochloric acid, first equivalence point appears at pH of about 8.3 and we know that phenolphtahein turns from colourless to fuchsia color at pH>8.2, hence phenolphthalein is used as an indicator for the first equivalence point whereas second equivalence point appears at pH of about 3.9 hence we use methyl orange as indicator for second equivalence point.

Titration of NH₄OH (weak base) with HCl (strong acid) solution results in acidic solution at equivalence point. Hence, methyl orange can be used as indicator which changes colour in pH range of 3.1 to 4.4.

For choosing an indicator in titration, we must note that pK (indicator) = pH of solution at equivalence point. pK value of bromothymol is 7.0 and pH value of solution at equivalence in titration of strong acid vs strong base is also 7.0, hence bromothymol is used as an indicator in reaction between strong base and strong acid.

This option is correct because , from n + l  rule . higher the value of this term higher is the energy . Now n value is 4 in this case . and l values are in the order of  s ( 0 ) < p ( 1 ) < d ( 2 ) < f ( 3 ) . Therefore, the correct order  is  s < p < d < f . Hence, this option is correct .

this option is correct because , energy is given by , E = -(13.6 Z² ) / n² . hence energy is dependent on principal quantum number . hence this option is correct .

This option is correct becausI radial node = n - l -1 . planar node =  l . total nodes = n - 1 . in this case n = 4, hence, total number nodes = 3 .