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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
151. |
For which orbital angular probability distribution is maximum at an angle of `45^@` to the axial direction ?A. `d_(x^2 - y^2)`B. `d_(z^2)`C. `d_(xy)`D. `P_x` |
Answer» Correct Answer - C ( c) The lobes of `d_(xy)` orbital are at an angle of `45^@` with `X` and `Y` axis. So along the lobes, angular probability distribution is maximum. |
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152. |
If n and `l` are respectively the principal and azimuthal quantum numbers , then the expression for calculating the total number of electrons in any energy level is :A. `sum_(l = 1)^(l = n) 2(2 l + 1)`B. `sum_(l = 1)^(l = n = 1) 2(2 l + 1)`C. `sum_(l = 0)^(l = n + 1) 2(2 l + 1)`D. `sum_(l = 0)^(l = n - 1) 2(2 l + 1)` |
Answer» Correct Answer - D (d) Total number of electrons in an orbital `= 2(2 l + 1)` The value of `l` varies from `0` to `n - 1`. `:.` Total number of electrons in any orbit `= sum_(l = 0)^(l = n + 1) 2 (2l + 1)`. |
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153. |
Match the column: ` |
Answer» Correct Answer - (A)Q,R(B)P,Q,R,S(C)P,S | |
154. |
In a hydrogen spectum if electron moves from `6^(th)` to `2^(nd)` by transition in multi steps then find out the number of lines in spectum |
Answer» Total number of line `=4+3+2+1+0=10` Total number of lines `=((n_(2)-n_(1))[(n_(2)-n_(1)+1)])/(2) = ((6-2)(4+1))/(2) implies (4xx5)/(2) = 10` |
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155. |
In a hydrogen spectum if electron moves from `7` to `1` orbit by transition in multi steps then find out the total number of lines in the spectum. |
Answer» `{:(,"Lyman",=,(n_(1)-1)=7-1=6),(,"Balmer",=,(n_(2)_2)=7-2=5),(,"Paschen",=,(n_(2)-3)=7-3=4),(,"Bracket",=,(n_(2)-4)=7-4=3),(,"Pfund",=,(n_(2)-5)=7-5=2),(,"Humphrey",=,(n_(2)-6)=7-6=1),(,"Total",=,21):}` Total number of lines can be calculatd as follows : Total number of lines `= ((n_(2)-n_(1))[(n_(2)-n_(1))+1])/(2) = ((7-1)(6+1))/(2)=(42)/(2)=21` |
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156. |
In Balmer series of `H` atom spectrum, which electronic transitions represents `3^(rd)` line? |
Answer» In Balmer series `{:(,3^(rd)"to"2^(nd),rarr,1"line"),(,4^(th)"to"2^(nd),rarr,2"line"),(,5^(th)"to"2^(nd),rarr,3"line"),(,"Infinite to" 2^(nd),rarr,"Last line or limiting line"):}` So, Ans is `5^(th)` to `2^(nd)` line " "`rarr " "3^(rd)` line |
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157. |
Match the column: |
Answer» Correct Answer - (A)Q,R(B)P,QR,S(C)P,S number of radial node`=(n-l-1)` number of angular node -l] |
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158. |
Calculate the radius ratio of `3^(rd) & 5^(th)` orbit of `He^(+)`. `r=0.529xx(n^(2))/(Z) Å` At. Number of of He `=2` |
Answer» `therefore " "r^(3) = 0.529xx((3)^(2))/(2)=0.529xx(9)/(2)` `r^(5) = 0.529xx((5)^(2))/(2)=0.529xx(25)/(2)` Therefore `(r_(3))/(r_(s))=(0.529xx (3)^(2)/(2))/(0.529xx(5)^(2)/(2))` `(r_(3))/(r_(s))=(9)/(25)` |
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159. |
Match the column: `{:(,"Column I", , "Column II"),((A)," Ratio of velocity of electron in " 5^(th)" and "3^(rd) " exicted level for H-atom is",(P)," "(4)/(1)" " ),((B),"Ratio of wavelenght of series limit of Balmer to Lyman series for a H-atom is ",(Q),""(2)/(3)""),((C),"Ratio of wavelenght of photon corresponding to "beta"-line of Lyman series and "gamma"- line of Paschen series for a H-atom is ",(R),""(1)/(4)""),((D)," Ratio of energy difference between "3^(rd)" and "1^(st)" orbit of H-atom amd "He^(+)"ion is",(S),""(3)/(32)""):}` |
Answer» Correct Answer - (A)P,Q,R(B)P,(C)S,(D)R (A) `Valpha(Z)/(n)` `(V_(6))/(V_(4))=(4)/(6)=(2)/(3)` (B) `((1)/(lambda))_(B)=R((1)/(2^(2))-(1)/(oo^(2)))` `((1)/(lambda_(1)))=R((1)/(1^(2))-(1)/(oo^(2)))` `1/(lambda_(B)/(1/lambda_(L)))=(1)/(4)lambda_(B)/(lambda_(L))=(4)/(1)` (C) `{:(beta-"line",("Second line")),((1/lambda)_(L)=,R(1/(1^(2))-(1)/(3^(2)))):}` `=(8R)/(9)` `{:(gamma-"line","(Third line)"),((1/lambda)_(P)=,R(1/(3^(2))-(1)/(6^(2)))):}` `=R((6^(2)-3^(2))/(6^(2)xx3^(2)))=(27R)/(324)=(3R)/(36)=(R)/(12)` `(((1)/(lambda)_(P)))/((1)/(lambda))_(C)=((R)/(12))/((8R)/(9))` `=(9)/(96)R=(3)/(32)R` (D) `Ealpha-Z^(2)/n^(2)` `(E_(3)-E_(1))_(H)/(therefore(E_(3)-E_(1))_(He^(+)))=(1)/(4)` |
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160. |
Match the column: ` |
Answer» Correct Answer - (A)R;(B)P;(C)Q | |
161. |
In a hydrogen spectrum if electron moves from `6^(th)` to `3^(rd)` orbit by transition in multi steps then find out the following steps : |
Answer» (a) Calculation of total number of lines : `=((n_(2)-n_(1))[(n_(2)-n_(1))+1])/(2)=((6-3)[(6-3)+1])/(2)=(3xx4)/(2)=6` (b) Calculation of number lines present in `U.V`. region. `e^(-)` moves from `6^(th)` to `3^(rd)` orbit in multisteps. For `U.V.` region, `e^(-)` should becomes into `1^(st)` shell. So the number of lines in `U.V`. region zero. (c ) Calculation of toal number of lines in visible region. For visible region, `e^(-)` should becomes into `2^(nd)` shell, so the number of lines in visible region zero. (d) Calculation of total number of lines in I.R. region. In I.R. region, Paschen, Bracket and Pfund series are present. Number of lines in Paschen series`= n_(2)-3` `= 6-3` `=3` Number of lines in Bracket series ` = n_(2)-4` `6-4` `=2` Number of lines in Pfund series ` = n_(2)-5` `=6-5` `=1` So total number of lines `= 3+2+1 =6` |
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162. |
What is the percentage of deterium to heavy water ?A. `20 %`B. `80 %`C. `60 %`D. `40 %` |
Answer» Deaterium in `20` parts of `D_(2)O = 4` parts Deuterium in `100 `parts of `D_(2)O = (1)/(2) xx 100 = 20%` |
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163. |
The degenerate orbitals have ………..value of quantum number(s)……….. |
Answer» Correct Answer - Same, n and l | |
164. |
Which of the following sets of orbitals may be degenerateA. `2s, 2p_x , 2p_y`B. `3s, 3p_x , 3p_(xy)`C. 1s,2s,3sD. `2p_x, 2p_y, 2p_z` |
Answer» Correct Answer - D `2p_x, 2p_y, 2p_z` sets of orbital is degenerate |
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165. |
d orbitals are fold degenerate and are speclled as ……………. |
Answer» `Five d_(x^(2),d_(y),d_(yz)d_(xz)` and ` d_(x^(2) - y^(2)` | |
166. |
Number of degenerate orbitals in a level of H-atom having ` E_n = (Rh)/9` :A. `3`B. `6`C. `9`D. `14` |
Answer» Correct Answer - C `E =- (Rh).n^2 =- (Rh)/( N)=3` ` 3rd` orbit of H-atom having ` 1s, 2s, 2p, 3s, 3p, 3d` orbitals . `:.` Degenerate orbitals ` =1 (3s ) + 3 (3p) = 5 (3d) =9`. |
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167. |
A `0.66 kg` ball is moving wih a speed of `100 m//s`. The associated wavelength will be.A. `6.6 xx 10^-32 m`B. `6.6 xx 10^-34 m`C. `1.0 xx 10^-35 m`D. `1.0 xx 10^-32 m` |
Answer» Correct Answer - C ( c) `lamda = (h)/(mv) = (6.6 xx 10^-34)/(0.66 xx 100) = 1 xx 10^-35 m`. |
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168. |
Uncertainty in the position of an electron mass ` (9. 1 xx 10^(31) kg)` moving with a velocity ` 300 ms^(-1)` accurate uptp ` 0.001 %` will be :A. `19.2 xx 10^(-2)m`B. `5.76 xx 10^(-2)m`C. `1.92 xx 10^(-2)m`D. `3.83 xx 10^(-2)m` |
Answer» Correct Answer - C `Deltax = (h)/(4pi m. DeltaV)` |
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169. |
The total number of atomic orbitals in fourth energy level of an atom is.A. 4B. 8C. 16D. 32 |
Answer» Correct Answer - C ( c) Number of atomic orbitals ` = n^2` Number of atomic orbitals in `4^(th)` energy shell `= 4^2 = 16`. |
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170. |
Uncertainty in the position of an electron mass ` (9. 1 xx 10^(31) kg)` moving with a velocity ` 300 ms^(-1)` accurate uptp ` 0.001 %` will be :A. `19.2 xx 10^(-2)` mB. `5.76 xx 10^(-2)` mC. 1.92 mD. `3.84 xx 10^(-2)` m |
Answer» Correct Answer - C |
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171. |
Uncertainty in the position of an electron mass ` (9. 1 xx 10^(31) kg)` moving with a velocity ` 300 ms^(-1)` accurate uptp ` 0.001 %` will be :A. `19.2 xx 10^-2 m`B. `5.76 xx 10^-2 m`C. `1.92 xx 10^-2 m`D. `3.84 xx 10^-2 m` |
Answer» Correct Answer - C ( c) `Delta x = (h)/(4 pi m Delta u)` =`(6.626 xx 10^-34 xx 100)/(4 xx 3.14 xx 9.1 xx 10^-31 xx 0.001 xx 300)` ltbrge`1.93 xx 10^-2 m`. |
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172. |
Uncertainty in the position of an electron mass ` (9. 1 xx 10^(31) kg)` moving with a velocity ` 300 ms^(-1)` accurate uptp ` 0.001 %` will be :A. `5.76xx10^(-2)m`B. `1.92xx10^(-2)m`C. `3.84xx10^(-2)m`D. `19.2xx10^(-2)m` |
Answer» Correct Answer - B | |
173. |
The uncertainty in the position of an electron `(mass = 9.1 xx 10^-28 g)` moving with a velocity of `3.0 xx 10^4 cm s^-1` accurate up to `0.001 %` will be (Use `(h)/(4 pi)` in the uncertainty expression, where `h = 6.626 xx 10^-27 erg - s`)A. `3.84 cm`B. `1.92 cm`C. `7.68 cm`D. `5.76cm` |
Answer» Correct Answer - B `Delta x = (h)/(4pimDeltav)` |
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174. |
The uncertainty in the position of an electron `(mass = 9.1 xx 10^-28 g)` moving with a velocity of `3.0 xx 10^4 cm s^-1` accurate up to `0.001 %` will be (Use `(h)/(4 pi)` in the uncertainty expression, where `h = 6.626 xx 10^-27 erg - s`)A. `1.92 cm`B. `7.68 cm`C. `5.76 cm`D. `3.84 cm` |
Answer» Correct Answer - A (a) `Delta p = m xx Delta v` `Delta p = 9.1 xx 10^-28 xx 3.0 xx 10^4 = (0.001)/(100)` `Delta P = 2.37 xx 10^-24` Hence `Delta x = (h)/(Delta p xx 4 pi)` =`(6.626 xx 10^-27)/(2.73 xx 10^-28 xx 4 xx 3.14)` `Delta x = 1.92 cm`. |
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175. |
The ratio of radius of 2nd and 3rd Bohr orbit isA. `3:2`B. `9:4`C. `2:3`D. `4:9` |
Answer» Correct Answer - D `(r_(1))/(r_(2)) =(n_(1)^(2))/(n_(2)^(2))` |
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176. |
The mass of the electrons `9.8 xx 10^(-28)` gram and uncertainty in the velocity equal to `2 xx 10^(-3) cm//sec`. The uncertainty in the position of an electron is `(h = 6.62 xx 10^(-27)` ergsec)A. `2.9 xx 10^(+2)cm`B. `2.9 xx 10^(-2)cm`C. `2.9 xx 10^(-12)cm^(-1)`D. `2.9 xx 10^(+12)cm^(-1)` |
Answer» Correct Answer - B `Deltax.m. Delta v = (h)/(4pi)` |
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177. |
A photon of 300 nm is absorbed by a gas and then emits two photons. One photon has a wavelength of 496 nm then the wavelength of second photon in nm is :A. 759B. 859C. 959D. 659 |
Answer» Correct Answer - A |
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178. |
The energy associatied with the first orbit in the hydrogen atom is ` - 2 17 xx 10^(18) "J atom" ^(-1)` . What is the energy associated with the fifth orbit ? |
Answer» ` E_1 =- 2. 17 xx 10^(-18) J` ` E_5 = E_1 /5^2 = (-2 . 17 xx 10^(-18))/(25) =- 8. 68 xx 10^(20) J`. |
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179. |
A photon of ` 300 nm` ios absorbed by a gas and then re-emits two photons . One re-emitte photon has wavelnght ` 4967 nm`, the wavelength fo second re-emitte photon is :A. ` 757`B. ` 857`C. ` 957`D. ` 657` |
Answer» Correct Answer - A `E_("photon absorbed") = E_1 + E_2 ("Energy releasesd") ` or `( hc)/( lambda) = ( hc)/( lambda_1) + ( hc)/( lambda_2)` or `(1)/ (lambda ) = 1/(lambda _(1)) + (1)/ ( lambda_(2))`. |
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180. |
What is the maximum number of emission lines when the excited electron of a H atom in `n = 6` drop to the ground state? |
Answer» Correct Answer - A The maximum numbner of lines `= (n (n - 1))/(2) = (6(6 - 1))/(2) = 15` |
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181. |
The energy associatied with the first orbit in the hydrogen atom is ` - 2.17 xx 10^(-18) "J atom" ^(-1)` . What is the energy associated with the fifth orbit ? |
Answer» Correct Answer - A::B::C `E_(n) = (E_(1))/(n^(2)):` `E_(3) = (-2.18 xx 10^(-18))/((5)^(2)) = J "atom" ^(-1) = - 8.72 xx 10^(-20) J "atom"^(-1)` |
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182. |
Suppose `10^-17 J` of energy is needed by the interior of human eye to see an object. How many photons of green light `(lambda = 550 nm)` are needed to generate this minimum amount of energy ?A. 14B. 28C. 39D. 42 |
Answer» Correct Answer - B `E = (nhc)/(lambda)` |
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183. |
Suppose `10^17 J` of energy is needed by the interior of human eye to see an object. How many photons of green light `(1 = 550 nm)` are needed to generate this minimum amount of energy ?A. 14B. 28C. 39D. 42 |
Answer» Correct Answer - B (b) Let the number of photons required be `n`. `(nhc)/(lamda) = 10^-17` `n = (10^-17 xx lamda)/(hc) = (10^-17 xx 550 xx 10^-9)/(6.626 xx 10^-34 xx 3 xx 10^8)` =`27.6 ~= 28 "photons"`. |
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184. |
Suppose `10^(-17)J` of light energy is needed by the interior of the human eye to see an object. How many photons of green light `(lambda=550nm)` are needed to generate this minimum amount of energy? |
Answer» Correct Answer - 28 photons `1.6xx10^(-19)J=1eV` `10^(-17)=(10^(-17))/(1.6xx10^(-19))eV=0.655xx10^(2)` `E=(nhc)/(lambda)" "0.625xx10^(2)=n(1240)/(550)` `2.77xx10=n` |
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185. |
Suppose ` 10^(-17) J` of light enrgy is needed by the interor of human eue to see an object . The photons fo green light ` * lambda = 55 0 nm)` needed to see to object are :A. ` 27`B. ` 28`C. ` 29`D. ` 30` |
Answer» Correct Answer - B `E_(Needed ) = 10^(-17) = n. (hc)/( lambda)` Where (n) is no. of photons needed fo enrgy ` ( hc)/( lambda) ` to provide enrgy fo ` 10^(-17) J`. |
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186. |
Calcuate the wavelength of a moving electron having ` 4 . 55 xx 10 ^(-25) J` of kinetic enrgy . |
Answer» Kinetic energy ` = 1/2 m u^2 = 4. 55 xx 10^(-25) J` ` :. u^2 = ( 2 xx 4.55 xx 10 ^(-25))/( 9. 108 xx 10^(-31))` ` :. U = 10 ^3 m sec^(-1)` ` :. lambda = (h)/( m u) = ( 6.626 xx 10^(-34))/( 9. 108 xx 10^(-31) xx 10^3)` ` = 7. 27 xx 10^(-7) meter `. |
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187. |
Nuclear radius is of the order of `10^(-13) cm ` white atomic radius is of `10^(-8)cm` .Assuming the nucleus and the to be spherical .What frection of an atom is occupted by nucleus ? |
Answer» `("Number volume")/("Atomic volume") = ("Number volume")/("Atomic volume") =((4)/(3) pi (10^(-13))^(3))/((4)/(3)pi(10^(-8))^(3)) = 10^(-15)` | |
188. |
Assuming a spherical shape for fluorine nucleus , calcuate the radius and the nuclear density of flourine nucleus of mass number 19. |
Answer» Correct Answer - `3.73xx10^(-13)` cm, `7.616=10^(13)g cm^(-3)` |
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189. |
Calculate the density of flatrine nucleus supposing that the shape of the nucleus is spherical and its radius is `5 xx 10^(-13)` (Mass of `f =19` ams) |
Answer» Mass of the nucleus of F atom `= 19 + 1.66 xx 10^(24)g (1 "amu" = 1.660 xx 10^(-24) g)` Volume of the nucleus of F atom `= (4)/(3) pi r^(3)` `= (4)/(3) xx 3.14 (5 xx 10^(-13))^(3) = 525 xx 10^(-39) cm^(3)` Density of the neucleus of F atom = `("Mass" )/("Volume") = (10 xx 1.66 xx 10^(24)g)/(525 xx 10^(-29) cm^(3)) = 6.0 xx 10^(13) g cm^(3)` |
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190. |
Assertion (A) : For `n = 3,l1` may be `0,1` and `2` and m may be `0, +- 1` and `0,+- 1` , and `+- 2` Reason (R ) : For each value of n, there are `0` to `(n - 1)` possible value of l for eachvalue of l , there are `0 to +-l` valie of mA. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
Answer» Correct Answer - A (a) For principal quantum number `n`. `l = 0 "to"(n - 1)` and `m = -1` to `l` including zero. |
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191. |
Supposing that the pauil exclesion principal is non-exitent .Which of the following is the most unaccepatable configuration of Li in the ground state ?A. `1s^(2) 2s^(1)`B. `1s^(3)`C. `1s^(1)2s^(2)`D. `1s^(1)2s^(1)2p^(1)` |
Answer» As a matter of fect, the configuration given in b ,c and d are wrong but the configuration given in d is the most unacceptable because there is one electron in each of the three orbital and according to the pauil exclassion principal ,a maximum of two electron can be put in an orbital. | |
192. |
Wavelength of radiation emitted when an electron jumps from a state A and C is `3000 "Å"` and it is `6000"Å"` when the electron jumps from state B to C . Wavelength of the radiation emitted when an electron jumps from state A to B will be :A. `2000 "Å"`B. `3000 "Å"`C. `4000"Å"`D. `6000"Å"` |
Answer» Correct Answer - D |
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193. |
Wavelength of radiation emitted when an electron jumps from a state A and C is `3000 "Å"` and it is `6000"Å"` when the electron jumps from state B to C . Wavelength of the radiation emitted when an electron jumps from state A to B will be :A. `2000 "Å"`B. `3000 "Å"`C. `4000 "Å"`D. `6000 "Å"` |
Answer» Correct Answer - B |
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194. |
A light is passed through a sample of a single elecrton excited spice which causes further excitation to some higher energy level and it emit a photon of energy 12.09 eV in back trabsition to some lower energy higher energy level .Then (i) Fing=d out the atomic number of specie if Debroglie wave lenght of electron in first excited state is 221.4 pm. (ii) Find out the value of higher energy level and lower energy level in back transition . |
Answer» Correct Answer - (i)`Z=3`(ii)`n_(1);n_(2)=9` (i) `lambda=((150)/(V))^(1//2)Å` `lambda^(2)=(150)/(V)xx10^(-20)` `V=(150)/(221.54)xx(10^(-20))/(10^(-24))=30.6` `therefore`(T.E) First excited state=-`13.6((Z^(2))/2^(2))``=-30.6 eV` `Z=3 Ans` (ii) `E_(n2)-E_(n1)`= `DeltaE=12.09=13.6xx9((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `n_(1)=3`, `n_(2)=9` |
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195. |
Atomic radil of flaorine and meon in Angstrom units are respectively given byA. `0.72,1,60`B. `1.60,1,60`C. `0.72,0,72`D. None of these |
Answer» Correct Answer - A Electronic configeration of florine and meon are `2,7 and 2 ,8` respectively |
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196. |
Assertion (A) : For `n = 3,l1` may be `0,1` and `2` and m may be `0, +- 1` and `0,+- 1` , and `+- 2` Reason (R ) : For each value of n, there are `0` to `(n - 1)` possible value of l for eachvalue of l , there are `0 to +-l` valie of mA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
Answer» Correct Answer - A l(azimathal quantum number) is always less than` n ,m_(s)` may have any value equal to l |
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197. |
For the energy levels in an atom , which of the following statement is //are correct ?A. There are serven principle electron energy levelsB. The second principal energy level cab be have four sub-shell energy level and contains a maximum of eight electronsC. The M energy level can have a maximum energy than the `3d` sub-energy levelD. The `4s` sub-energy level is at a lower energy than the `3d` sub-energy level |
Answer» Statement (a) and (d) are true statement (b) is wrong because for `n = 2,l = 0,1` (two sub-energy levels) c is wrong because M shell means `n = 3` Maximum electrons d can have `= 2n^(2) = 2 xx 3^(2) = 18` Hence a and d are correct |
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198. |
Which of the following strtement is /are correct ?A. The electeron configuration of `Cr` is `[Ar]3d^(5) 4s^(1) ("atomic number of" Cr = 24)`B. The magnitic quantum number may have a negative valueC. In silver atom `23` electron have spin of one type and `24` of the opposite type (atomic number of `Ag = 47)`D. The oxidation state of nitrogen in `HN_(3)` is `-3` |
Answer» Only d is wrong because the oxidation state of N in `HN_(3) `is `- 1//3` Hence a,b, and c are correct |
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199. |
The ratio of the energy of a photon of wavelength `3000 "Å" ` to that of photon of wavelength `6000 "Å"`is :A. `(1)/(2)`B. 2C. 3D. `(1)/(3)` |
Answer» Correct Answer - B |
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200. |
Assertion (A) : The atomic radil of the electrons of oxygen family are smaller than the atomic radil of corresponding electrons of the nitrogen family Reason (R ) : The members of oxygen family are all more electromagative and thus have lower value of nuclear tahn those of the nitrogen famolyA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
Answer» Correct Answer - C Oxygen family has higher effective nuclear charge Hence , its atomic radil is smaller than nitrogen family . |
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