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InterviewSolution
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.
| 951. |
Which species has the maximum number of lone pair of electrons on the central atom ?A. `ClO_(3)^(-)`B. `XeF_(4)`C. `SF_(4)`D. `I_(3)^(-)` |
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Answer» Correct Answer - d |
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| 952. |
Which of the following species is non polar with presence of polar bond and lone pair of electron on central atom.A. `CO_(2)`B. `SF_(4)`C. `XeF_(4)`D. `CF_(4)` |
| Answer» Correct Answer - C | |
| 953. |
Which of the following pairs of species have the same bond order ?A. `CN^(-)` and `CN^(+)`B. `O_(2)^(-)` and `CN^(-)`C. `NO_(+)` and `CN^(+)`D. `CN^(-)` and `NO_(2)^(+)` |
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Answer» Correct Answer - D `NO^(+) =14e^(-)` `sigma1s^(2),sigma1s^(**2),sigma2s^(2),sigma2s^(**2),sigma2p_(z)^(2),|[pi 2p_(x)^(2)],[pi2 p_(y)^(2)]|` `B.O=1/2(10-4)=3` `CN^(-)=14e^(-)B.O.=3` `CN^(+)=12e^(-) B.O. =2` `O_(2)^(-)=17e^(-) B.O. =1.5` |
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| 954. |
Assertion(A) - Fluorine molecule has bond order one. Reason(R)-The number of electrons in antibonding molecular orbital is two less than that of bonding molecular orbitals.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. if both assertion and reason are true but reason is not the correct explanation of the assertionC. if assertion is true but reason is falseD. if the assertion and reason both are false |
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Answer» Correct Answer - A MO electronic configuration of `F_(2)` molecule is `F_(2)=sigma1s^(2),sigma^(**)1s^(2),sigma2s^(2),sigma^(**)2s^(2),sigma2p_(x)^(2).` `pi 2 p_(y)^(2)=pi 2p_(z)^(2), pi^(**) 2p_(y)^(2)=pi^(**)2p_(z)^(2)` Bond order `=(N_(b)-N_(a))/2` Where, `N_(b)`=Number of electrons in bonding molecular orbitals `N_(a)`=Number of electrons in antibonding molecular orbitals `=(8-6)/2=1` |
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| 955. |
Which of the following pairs of species have the same bond orderA. `O_(2),NO^(+)`B. `CN^(-),CO`C. `N_(2),O_(2)^(-)`D. CO,NO |
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Answer» Correct Answer - B Total no. of electrons in `CN^(-)` is 14 Total no. of electrons in CO is also 14 Hence , B.O. of both `CN^(-)` and CO is 3 |
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| 956. |
Assertion `B_(2)` molecule is diamagnetic Reasoning The highest occupied molecular orbital is of sigma type .A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. if both assertion and reason are true but reason is not the correct explanation of the assertionC. if assertion is true but reason is falseD. if the assertion and reason both are false |
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Answer» Correct Answer - D In `B_(2)`, total number of electrons =10 `B_(2) to sigma(1s)^(2)sigma^(**)(1s)^(2)sigma(2s)^(2)sigma^(**)(2s)^(2)sigma(2p_(x))^(1)pi(2p_(y))^(1)` Presence of unpaired electrons show the paramagnetic nature. The highest occupeid molecular orbital is of `pi` -type |
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| 957. |
Estimate the lattice energy of `CaCO_3` if `r_(ca^+) =114 pm` and `r_(CO_3^(2-) =185 p m`. |
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Answer» By usign Kapustinskil equation , `U =- 121,000 ((vz_1z_2)/d) ( - ( 34.5)/d)` kJ //mol `( v =2, z_2 =-2 , d= gamma^+ =gamma^-` ` =114 + 185 =299 p m)` `U8 =- 121,000 xx ((2xx2xx(-2))/(299)) (1- ( 32 .5)/(299))` `=2860` kJ //mol. |
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| 958. |
The smallest bond angle around the central atom will be there inA. `IF_(7)`B. `CH_(4)`C. `BeF_(2)`D. `BF_(3)` |
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Answer» Correct Answer - A It is shows `sp^(3)d^(3)`-hybridization. Hence the bond angle is about `72^(@)` |
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| 959. |
If the molecule of HCl were totally polar, the expected value of dipole moment is 6-12 D (debye) but the experimental value of dipole moment was 1.03 D. Calculate the percentage ionic character.A. 17B. 83C. 50D. Zero |
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Answer» Correct Answer - A Percent ionic character `=("Obs. dipole moment")/("Cal. dipole moment")xx100` `=(1.03D)/(6.12D)xx100=17%` |
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| 960. |
Which one of the following molecules has the smallest bond angle ?A. `NH_3`B. `PH_3`C. `H_2O`D. `H_2Se` |
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Answer» Correct Answer - D All these molecules involved `sp^3`-hybridisation with one lone pair (`NH_3 and PH_3`) or two lone pairs (`H_2O`, `H_2Se` and `H_2S`). More the number of lone pairs, smaller is the angle. Out of `H_2O`, `H_2S` and `H_2Se`, H-Se-H angle in `H_2Se` is least (`91^(@)`). |
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| 961. |
In which of the following pairs, the two species are iso-structural ? `{:((a)SO_(4)^(2-)and NO_(3)^(-),(b),BF_(3)andNF_(3)),((c )BrO_(3)^(-)and XeO_(3),(d),SF_(4)and XeF_(4)):}`A. `BrO_3^(-) and XeO_3`B. `SF_4 and XeF_4`C. `SO_3^(2-) and NO_3^-`D. `BF_3 and NF_3` |
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Answer» Correct Answer - A `BrO_3^(-) and XeO_3` both are pyramidal |
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| 962. |
Which element out of L, M, Q. R will most readily form a diatomic molecule ?A. `L:[He] 2s^2 2p^3`B. `M:[He]2s^2 2p^5`C. `Q:[Ne]3s^1`D. `R:[Ne] 3s^2 3p^2` |
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Answer» Correct Answer - B The M-atom requiring 1 electron to complete its octet forms a mono covalent bond with another M-atom |
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| 963. |
`NH_(3)and BF_(3)`from adduct readily because they fromA. ionic bondB. covalent bondC. co-ordinate bondD. hydrogen bond. |
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Answer» Correct Answer - C `NH_3` acts as lewis base while `BF_3` acts as lewis acid `H_3N : to BF_3` |
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| 964. |
Which of the following species are planar ?A. `CO_(3)^(2-)`B. `NH_(3)`C. `PCl_(3)`D. None of these |
| Answer» Correct Answer - A | |
| 965. |
Which of the following does not involves `dsp^(2)`-hybridisation and are square planar ?A. `NiCl_(4)^(2-)`B. `SCl_(4)`C. `NH_(4)^(+)`D. `PtCl_4^(2-)` |
| Answer» Correct Answer - D | |
| 966. |
Which of the following molecule is planar?A. `CH_(4)`B. `NH_(3)`C. `C_(2)H_(4)`D. `SiCl_(4)` |
| Answer» Correct Answer - C | |
| 967. |
Which one of the following molecules is planar?A. `NF_(3)`B. `NCl_(3)`C. `PH_(3)`D. `BF_(3)` |
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Answer» Correct Answer - d |
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| 968. |
Among the following species, identify the isostuctural pairs `NF_(3). NO_(3)^(-), BF_(3), H_(3)O, HN_(3)`A. `[NF_(3),NO_(3)^(-)]` and `[BF_(3),H_(3)O^(+)]`B. `[NF_(3),N_(3)H]` and `[NO_(3)^(-),BF_(3)]`C. `[NF_(3),H_(3)O^(+)]` and `[NO_(3)^(-),BF_(3)]`D. `[NF_(3),H_(3)O^(+)]` and `[N_(3)H,BF_(3)]` |
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Answer» Correct Answer - c |
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| 969. |
Among the following species, identify the isostuctural pairs `NF_(3). NO_(3)^(-), BF_(3), H_(3)O, HN_(3)`A. `[NF_(3),NO_(3)^(-)]` and `[BF_(3),H_(3)O^(+)]`B. `[NF_(3)HN_(3)]` and `[NO_(3)^(-),BF_(3)]`C. `[NF_(3),H_(3)O^(+)]` and `[NO_(3)^(-),BF_(3)]`D. `[NF_(3),H_(3)O^(+)]` and `[HN_(3),BF_(3)]` |
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Answer» Correct Answer - C Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus, In `NF_(3): H=1/2 (5+3-0+0)=4` Thus,N in `NF_(3)` is `sp^(3)` hybridized as 4 orbitals are involved in bonding. In `NO_(3)^(-): H=1/2(5+3-0+0)=3` Thus N in `NO_(3)^(-)` is `sp^(2)` hybridized as 3 orbitals are involved in bonding. in `BF_(3): H=1/2(3+3-0+0)=3` Thus B in `BF_(3)` is `sp^(2)` hybridized as again 3 orbitals are involved in bonding In `H_(3)O^(+): H=1/2(6+3-1+0)=4` Thus, O in `H_(2)O^(+)` is `sp^(3)` hybridized as 4 orbitals are involved in bonding. Thus, isostructural pairs are `[NF_(3),H_(3)O^(+)]` and `[NO_(3)^(-),BF_(3)]`. |
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| 970. |
According to the electronic theory of chemical bonding, developed independently by `Kӫssel` and Lewis, the atoms of representative elements can combine either by the transfer of valence electrons from one atom to another (gaining or losing) or by the sharing of valence electrons in order to have an octet in their ______________. This is known as octet rule.A. inner shellB. penultimate hsellC. antepenultimate shellD. outermost shell |
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Answer» Correct Answer - D When atoms interact to from a chemical bond, only their outermost regions are in contact. Outermost shell electrons of representative elements are called valency electrons because they decide the common as well as the variable valence. |
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| 971. |
Indicate the nature of bonding in `C Cl_(4)` and `CaH_(2)`A. Covalent in `C Cl_(4)` and electrovalent in `CaH_(2)`B. Electrovalent in both `C Cl_(4)` and `CaH_(2)`C. Covalent in both `C Cl_(4)` and `CaH_(2)`D. Electrovalent in `C Cl_(4)` and covalent in `CaH_(2)` |
| Answer» Correct Answer - A | |
| 972. |
The nature of chemical bonding in graphite isA. CovalentB. ionicC. MetallicD. Coordinate |
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Answer» Correct Answer - A In graphite all carbon atoms are `sp^(2)`-hybridised and have covalent bond. |
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| 973. |
The molecules which contain `sigma` bond , `pi` bond and lone pair `1:1:1` ratio :-A. `C_(2)N_(2)`(Cyanogen )B. `C_(6)H_(6)` ( Benzene )C. `C(CN)_(4)` (Tetracyano methane )D. `C_(3)O_(2)` ( Carbon suboxide ) |
| Answer» Correct Answer - D | |
| 974. |
The species that does not contain peroxied bond is //are :A. `PbO_2`B. `H_2O_(2)`C. `MnO_(2)`D. `BaO_(2)` |
| Answer» Correct Answer - A::C | |
| 975. |
Which one of the following constitutes a group of the isoelectronic species ?A. `CN^- , N_2 , O_2^(2-), C_2^(2-)`B. `N_2, O_2^(2-), NO^+ , CO`C. `C_2^(2-), O_2^(-), CO,NO`D. `NO^(+), C_2^(2-), CN^(-), N_2` |
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Answer» Correct Answer - D `CN^(-)` =6+7+1= 14 `e^-` `N_2` =8+8+2= 14 `e^-` `O_2^-` =8+8+2= 16 `e^-` `C_2^(2-)` =6+6+2= 14 `e^-` `O_2^(-)` =8+8+1= 17 `e^-` `NO^(+)` =7+8-1= 14 `e^-` CO =6+8= 14 `e^-` |
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| 976. |
Which one of the following constitutes a group of the isoelectronic species ?A. `C_2^(2-), O_2^(-), CO, NO`B. `N_2, O_2^(-), NO^(+), CO`C. `CN^(-), N_2, O_2^(2-), C_2^(2-)`D. `NO^(+), C_2^(2-), CN^(-), N_2` |
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Answer» Correct Answer - D Isoelectronic species have the same electronic configuration due to the same number of electrons. `C_2^(2-)=6+6+2=14, O_2^(-)=8+8+1=17` `CO=6+8=14, NO=7+8=15, N_2=7+7=14` `NO^(+)=7+8-1=14, CN^(-)=6+7+1=14` |
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| 977. |
Which one of the following has the highest dipole moment ?A. `AsH_3`B. `SbH_3`C. `PH_3`D. `NH_3` |
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Answer» Correct Answer - D In the given molecules nitrogen has greater electronegativity .So, it has greater dipole moment and correct order of dipole moment is `NH_3 gt PH_3 gt AsH_3 gt SbH_3` |
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| 978. |
Which of the following molecule is nonpolar? (i) `PbCl_4` (ii) `BF_3` (iii) `SnCl_2` (iv) `CS_2`A. (i), (ii), (iii)B. (i), (ii), (iii), (iv)C. (i), (ii), (iv)D. (ii), (iii), (iv) |
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Answer» Correct Answer - C Molecules with symmetrical shapes have zero dipole moments. `PbCl_4` is regular tetrahedron, `BF_3` is trigonal planar, while `CS_2` is linear. All have zero dipole moment and, hence, are nonpolar. On the other hand, `SnCl_2` is a bent molecule. Thus, its bond dipoles do not cancel and `SnCl_2` is a polar molecule. |
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| 979. |
Dipole moment is highest for:A. `CHCl_3`B. `CH_4`C. `CHF_3`D. `C-Cl_4` |
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Answer» Correct Answer - C ` C-F` bnond is more polar. |
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| 980. |
`N_2` and `O_2` are converted to mono cations `N_2^+` and `O_2^+` respectively, which statement is wrong ?A. In `N_2^+`, the N-N bond weakensB. In `O_2^+`, the O-O bond order increasesC. In `O_2^+`, paramagnetism decreasesD. `N_2^+` becomes diamagnetic. |
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Answer» Correct Answer - D In `N_2^(+)`, there is one unpaired electron. Hence it is paramagnetic. |
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| 981. |
Which bond angel ` theta ` would result in the maximum dipole moment for the triatomic molecule `XY_2` shown below ? ` A. ` theta = 90^@`B. ` theta = 120^@`C. ` theta =150^@`D. ` theta =180^@` |
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Answer» Correct Answer - A ` mu =sqrt(mu_1^2 +mu_2^2 + 2mu_1mu_2 cos theta , ) if `theta = 90^@`. `mu` is maximum . |
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| 982. |
Which bond angle `theta` would result in the maximum dipole moment for the triatomic `YXY`?A. `theta=90^(@)`B. `theta=120^(@)`C. `theta=150^(@)`D. `theta=180^(@)` |
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Answer» (a) `mu=sqrt(mu_(1)^(2)+mu_(2)^(2)+2mu_(2)mu_(2) cos theta )` If `theta=90^(@)`, `mu` is maximum. Because `cos90^(@)`=0, since, the angle increases from 90to 180, the value of `cos theta` becomes more and more negative and hence resultant decreses. Thus, dipole moment is maximum, when `theta=90^(@)`. |
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| 983. |
What is true about `CN^-` and `N_2` ?A. Both are isoelectronicB. Both are chemically inertC. Both are highly reactiveD. Both have same polarity of bonds. |
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Answer» Correct Answer - A Both `CN^-` and `N_2` have 14 electrons and are therefore isoelectronic. |
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| 984. |
The correct order of increasing bond angles in the following triatomic species isA. `NO_2^(-) lt NO_2^(+) lt NO_2`B. `NO_2^(-) lt NO_2 lt NO_2^(+)`C. `NO_2^(+) lt NO_2 lt NO_2^(-)`D. `NO_2^(+) lt NO_2^(-) lt NO_2` |
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Answer» Correct Answer - B According to VSEPR theory, The order of repulsion of electrons is lp-lp repulsion gt lp-bp repulsion gt bp-bp repulsion Higher the number of lone pair of electrons (greater is repulsion ) lower is the bond angle and vice-versa. In `NO_2^-` one lone pair of electron is present while in `NO_2^+` no lone pair of electron is present. Hence the correct order of bond angles is `NO_2^(-) lt NO_2 lt NO_2^+` |
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| 985. |
The species having bond order different from that in `CO` isA. `NO^-`B. `NO^+`C. `CN^-`D. `N_2` |
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Answer» Correct Answer - A `NO^+, CN^(-)` and `N_2` are isoelectronic with CO (containing 14 electrons each) Hence all of them have the same bond order , viz, 3 . `NO^-` has 16 electrons with the same configurations as that of `O_2`. Hence its bond order = 2. |
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| 986. |
Four diatomic species are listed in different sequence .Which of these represent the correct order of their increasing bond order?A. `O_2^(-) lt NO lt C_2^(2-) lt He_2^+`B. `NO lt C_2^(2-) lt O_2^(-) lt He_2^(+)`C. `C_2^(2-) lt He_2^(+) lt NO lt O_2^(-)`D. `He_2^(+) lt O_2^(-) lt NO lt C_2^(2-)` |
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Answer» Correct Answer - D The molecular orbital configuration of `O_2^-` (8+8+1=17)=`sigma1s^2,overset***sigma1s^2,sigma2s^2,overset***sigma2s^2,sigma2p_x^2,pi2p_y^2~~pi2p_z^2,overset***2p_y^2~~overset***pi2p_z^1` Bond order (BO)=`(N_b-N_a)/2=(10-7)/2=1.5` `NO` (7+8=15)=`sigma1s^2,overset***sigma1s^2,sigma2s^2,overset***sigma2s^2,sigma2p_x^2,pi2p_y^1~~pi2p_z^0` Bond order (BO)=`(10-5)/2=2.5` `C_2^(2-)` (6+6+2=14)=`sigma1s^2,overset***sigma1s^2,sigma2s^2,overset***sigma2s^2,pi2p_y^2~~pi2p_z^2,overset***2p_x^2` Bond order (BO)=`(10-4)/2=3` `He_2^+` (2+2-1=3) =`sigma1s^2, overset***1s^1` BO`=(2-1)/2=1/2=0.5` Hence, order of increasing bond order is `He_2^(+) lt O_2^(-) lt NO lt C_2^(2-)` |
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| 987. |
The common features among the species ` CN^- , CO ` and ` NO^+` are :A. Bond order three and isoelectronicB. Bond order three and weak field ligandsC. Bond order three and `pi`-acceptorsD. Isoelectronic and weak field ligands. |
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Answer» Correct Answer - A All the species are soelectronic because each one of them has `14e^(-)` MO configuration is `KK(sigma_(2s)^**)^2 (sigma_(2s)^**)^2 (pi_(2p_z))^2 (pi_(2p_x))^2 (pi_(2p_y))^2` Bond order is `1/2` (10-4) =3 |
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| 988. |
The nodal plane is the pi -bond of ethene is located in :A. the molecular planeB. a plane parallel to moelcular planeC. a plane perpendicular to the molecular plane which bisects the carbon -carbon sigma bond at right anglesD. a plane perpendicular to the moelcular plane which contains the carbon-carbon sigma bond |
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Answer» Correct Answer - A A `pi`-bond has nodel plane passing through the two bonded nuclei i.e., molecular plane . |
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| 989. |
The number of nodal planes present in a `sigma^**` antibonding orbital is |
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Answer» Correct Answer - C `sigma_(s)^(**)` Antibonding orbitals have only one nodal plane |
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| 990. |
Four diatomic species are listed in different sequence .Which of these represent the correct order of their increasing bond order?A. `O_2^(-) lt NO lt C_2^(2-) lt He_2^(+)`B. `NO lt C_2^(2-) lt O_2^(-) lt He_2^(+)`C. `C_2^(2-) lt He_2^(+) lt NO lt O_2^(-)`D. `He_2^(+) lt O_2^(-) lt NO lt C_2^(2-)` |
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Answer» Correct Answer - D Electronic configuration are `He_2^(2+): (sigma1s)^2(sigma^**1s)^1`, `bo=(2-1)/(2)=1/2=0.5` `O_2^(-):KK(sigma2s)^2(sigma^**2s)^2(sigma2p_z)^2(pi2p_x^2=pi2p_y^2)(pi^**2p_x^2=pi^**2p_y^1)` `bo=(10-7)/(2)=1.5` `NO:KK(sigma2s)^2(sigma^**2s)^2(sigma2p_z)^2(pi2p_x^2=pi2p_y^2)(pi^**2p_x^1=pi^**2p_y^0)` `bo=(10-5)/(2)=2.5` `C_2^(2-):KK(sigma2s)^2(sigma^**2s)^2(pi2p_x^2=pi2p_y^2)(sigma2p_z)^2` `bo=(10-4)/(2)=3.0` Thus, the correct order of increasing bond order is `He_2^(+) lt O_2^(-) lt NO lt C_2^(2-)` |
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| 991. |
The nodal plane in the `pi`-bond of ethene is located in:A. the molecular planeB. a plane parallel to the molecular planeC. a plane perpendicular to the molecular plane which bisects the carbon-carbon sigma bond at right anglesD. a plane perpendicular to the molecular plane which containsthe carbon-carbon `sigma` bond. |
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Answer» Correct Answer - A The molecular plane does not have any `pi`-electron density |
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| 992. |
Which of the following molecular species has unpaired electrons(s) ? .A. `N_2`B. `F_2`C. `O_2^-`D. `O_2^(2-)` |
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Answer» Correct Answer - C `O_2^(-)` has one unpaired electron in `pi^(**)(2p)` orbital. |
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| 993. |
Which of the following hydrocarbons has the lowest dipole moment?A. B. `CH_3C-=C CH_3`C. `CH_3CH_2C=CH`D. `CH_2=CH-C-=CH` |
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Answer» Correct Answer - B (B) is symmetrical with linear structure and will therefore, have zero dipole moment. |
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| 994. |
Which of the following are Lewis acids?A. `AlCl_(3)` and `SiCl_(4)`B. `PH_(3)` and `SiCl_(4)`C. `BCl_(3)` and `AlCl_(3)`D. `PH_(3)` and `BCl_(3)` |
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Answer» Correct Answer - C `BCl_(3)` and `AlCl_(3)` both have incomplete octate and act as lewis acid |
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| 995. |
The structural formula of a compound is `CH_(3)-CH=C=CH_(2)`. The type of hybridization at the four carbons from left to right areA. `sp^(2),sp,sp^(2),sp^(3)`B. `sp^(2),sp^(3),sp^(2),sp`C. `sp^(3),sp^(2),sp,sp^(2)`D. `sp^(3),sp^(2),sp^(2)sp^(2)` |
| Answer» Correct Answer - C | |
| 996. |
The hydrogen bonds are encountered in `HF, H_2 O, NH_3` and `HF_2^-`. The relative order of energies of hydrogen bonds isA. `HF gt H_(2)O gt H_(3)N gt HF_(2)^(-)`B. `H_(2)O gt HF_(2)^(-) gt HF gt NH_(3)`C. `HF gt HF_(2)^(-) gt H_(2)O gt NH_(3)`D. `HF_(2)^(-) gt HF gt H_(2)O gt NH_(3)` |
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Answer» Correct Answer - D Higher is the electronegativity, stronger is the H-bond. Fluorine has more electronegativity among the given compound. In `HF_(2)^(-)` there are two fluorine atoms that results into even stronger H-bonding. |
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| 997. |
How many `pi`-bonds are in `H_2 S_2 O_6` ? |
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Answer» `H_2 S_2 O_6` Has ` HO-underset(O)underset(||)overset(O)overset(||)(S)-underset(O) underset(||)overset(O)overset(||)(S)-OH` pi - bonds `=4`. |
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| 998. |
Find the formal charge of the O-atoms in ` [overset (..) ( :O) = N = overset (..)(O: )^+ ]`ion . |
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Answer» Formal cahrge = Total valence electron in free atom -No . of non-bonding electrons ` -1 /2 xx NO. ` of bonding electrons ` = 6 -4 - 1/2 xx 4 =0` . |
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| 999. |
In the cyanide ion, the formal negative charge is on :A. ` C`B. ` N`C. both C and ND. resonate between C and N |
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Answer» Correct Answer - D Cyandie ion is, `-C equiv N lt lt - bar(N) equiv C` |
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| 1000. |
The correct order of bond angles in the molecules, `H_2O`, `NH_3`, `CH_4`, and `CO_2` isA. `H_2O gt NH_3 gt CH_4 gt CO_2`B. `H_2O lt NH_3 lt CO_2 lt CH_4`C. `H_2O lt NH_3 gt CO_2 gt CH_4`D. `CO_2 gt CH_4 gt NH_3 gt H_2O` |
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Answer» Correct Answer - D `CO_2` is linear, `CH_4` is tetrahedral with zero lone pair, `NH_3` is tetrahedral with one lone pair, while `H_2O` is tetrahedral with two lone pairs. Thus, the order of bond angle is `CO_2(180^@)gtCH_4(109.50^@)gtNH_3(107^@)gtH_2O(104.5^@)` |
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