Which of the following have no unit?
Question: Which of the following have no unit? (i) Electronegativity (ii) Electron gain enthalpy (iii) Ionisation enthalpy (iv) Metallic character Solution: Option (i) and (iv) are the answers....
Read More →In which of the following options order
Question: In which of the following options order of arrangement does not agree with the variation of property indicated against it? (i) Al3+ Mg2+ Na+ F(increasing ionic size) (ii) B C N O (increasing first ionisation enthalpy) (iii) I Br Cl F (increasing electron gain enthalpy) (iv) Li Na K Rb (increasing metallic radius) Solution: Option (ii) and (iiii) are the answers....
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $x=a t^{2}, y=2$ at at $t=1$ Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to $t$ $\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{at}$ $\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{a}$ Now dividing $\frac{\mathrm{dy}}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ to obtain the slope of tangent $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}}$ $\mathrm...
Read More →Which of the following sets
Question: Which of the following sets contain only isoelectronic ions? (i) Zn2+, Ca2+, Ga3+, Al3+ (ii) K+, Ca2+, Sc3+, Cl (iii) P3-, S2-, Cl, K+ (iv) Ti4+, Ar, Cr3+, V5+ Solution: Option (ii) K+, Ca2+, Sc3+, Cl and (iii)P3-, S2-, Cl, K+are the answers....
Read More →Which of the following statements are correct?
Question: Which of the following statements are correct? (i) Helium has the highest first ionisation enthalpy in the periodic table. (ii) Chlorine has less negative electron gain enthalpy than fluorine. (iii) Mercury and bromine are liquids at room temperature. (iv) In any period, the atomic radius of alkali metal is the highest. Solution: Option (i), (iii) and (iv) are the answers....
Read More →Which of the following elements
Question: Which of the following elements will gain one electron more readily in comparison to other elements of their group? (i) S (g) (ii) Na (g) (iii) O (g) (iv) Cl (g) Solution: Option (i)S (g)and (iv)Cl (g) are the answers....
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $\mathrm{x}=\frac{2 \mathrm{at}^{2}}{1+\mathrm{t}^{2}}, \mathrm{y}=\frac{2 \mathrm{at}^{3}}{1+\mathrm{t}^{2}}$ at $\mathrm{t}=1 / 2$ Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to $t$ $\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\left(1+\mathrm{t}^{2}\right) 4 \mathrm{at}-2 \mathrm{at}^{2}(2 \mathrm{t})}{\left(1+\mathrm{t}^{2}\right)^{2}}$ $\frac{\mathrm{dx}...
Read More →Which of the following sequences
Question: Which of the following sequences contain atomic numbers of only representative elements? (i) 3, 33, 53, 87 (ii) 2, 10, 22, 36 (iii) 7, 17, 25, 37, 48 (iv) 9, 35, 51, 88 Solution: Option (i) and (iv) are the answers....
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{(b+c)}{a} \cdot \cos \frac{(B+C)}{2}=\cos \frac{(B-C)}{2}$ Solution: Need to prove: $\frac{(b+c)}{a} \cdot \cos \frac{(B+C)}{2}=\cos \frac{(B-C)}{2}$ We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius. Therefore, $a=2 R \sin A \cdots$ (a) Similarly, $b=2 R \sin B$ and $c=2 R \sin C$ Now, $\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{2 \mathrm{R} \sin \mathrm{A}}{2 \mathrm{R} \sin \mathrm{B}+2 \mathrm{R} ...
Read More →Those elements impart colour to the flame
Question: Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionisation (i.e., absorb energy in the visible region of the spectrum). The elements of which of the following groups will impart colour to the flame? (i) 2 (ii) 13 (iii) 1 (iv) 17 Solution: Option (i) and (iii) are the answers....
Read More →Which of the following elements
Question: Which of the following elements can show covalency greater than 4? (i) Be (ii) P (iii) S (iv) B Solution: Option (ii) Pand (iii)S are the answers....
Read More →Electronic configurations of four elements A, B, C
Question: Electronic configurations of four elements A, B, C and D are given below : (A) 1s22s22p6 (B) 1s22s22p4 (C) 1s22s22p63s1 (D) 1s22s22p5 Which of the following is the correct order of increasing tendency to gain electron : (i) A C B D (ii) A B C D (iii) D B C A (iv) D A B C Solution: Option (i) is the answer....
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{(a+b)}{c} \sin \frac{C}{2}=\cos \frac{(A-B)}{2}$ Solution: Need to prove: $\frac{(\mathrm{a}+\mathrm{b})}{\mathrm{c}} \sin \frac{\mathrm{C}}{2}=\cos \frac{(\mathrm{A}-\mathrm{B})}{2}$ We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius. Therefore, $a=2 R \sin A \cdots$ (a) Similarly, $b=2 R \sin B$ and $c=2 R \sin C$ Now, $\frac{a+b}{c}=\frac{2 R(\sin A+\sin B)}{2 R \sin C}=\frac{\sin A+\sin B}{\sin C}$ The...
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\pi / 2$ Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to theta $\frac{\mathrm{dx}}{\mathrm{d} \theta}=1+\cos \theta$ $\frac{d y}{d \theta}=-\sin \theta$ Dividing both the above equations $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\sin \theta}{1+\cos \theta}$ $\mathrm{m}$ (tangent) at theta $(\pi / 2)=-1$ n...
Read More →Comprehension given below is followed by
Question: Comprehension given below is followed by some multiple-choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which are related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic the table has been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and ...
Read More →The formation of the oxide ion,
Question: The formation of the oxide ion, O2 (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below: O (g) + e O(g) ; ; ∆ HV = 141 kJ mol1 O (g) + e O2 (g) ; ∆ HV = + 780 kJ mol1 Thus the process of formation of O2 in the gas phase is unfavourable even though O2 is isoelectronic with neon. It is due to the fact that (i) oxygen is more electronegative. (ii) addition of electron in oxygen results in larger size of the ion. (iii) electron repulsion outweighs ...
Read More →Which of the following is the correct
Question: Which of the following is the correct order of the size of the given species: (i) I I I+ (ii) I+ I I (iii) I I+ I- (iv) I I I+ Solution: Option (iv) I I I+is the answer....
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{(b-c)}{a} \cos \frac{A}{2}=\sin \frac{(B-C)}{2}$ Solution: Need to prove: $\frac{(\mathrm{b}-\mathrm{c})}{\mathrm{a}} \cos \frac{\mathrm{A}}{2}=\sin \frac{(\mathrm{B}-\mathrm{C})}{2}$ We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius. Therefore, a = 2R sinA ---- (a) Similarly, b = 2R sinB and c = 2R sinC From Left hand side, $=\frac{2 R \sin B-2 R \sin C}{2 R \sin A} \cos \frac{A}{2}$ $=\frac{2 \cos \left...
Read More →Find the equation
Question: Find the equation of the tangent to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\pi / 4$. Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to theta $\frac{\mathrm{dx}}{\mathrm{d} \theta}=1+\cos \theta$ $\frac{\mathrm{dy}}{\mathrm{d} \theta}=-\sin \theta$ Dividing both the above equations $\frac{d y}{d x}=-\frac{\sin \theta}{1+\cos \theta}$ $m$ at theta $(\pi / 4)=-1+\frac{1}{\sqrt{2}}$ equation of tangent is given by $y-y_{1}=m(\operat...
Read More →The elements in which electrons are
Question: The elements in which electrons are progressively filled in 4f-orbital are called (i) actinoids (ii) transition elements (iii) lanthanoids (iv) halogens Solution: Option (iii) lanthanoidsis the answer....
Read More →The period number in the long form
Question: The period number in the long form of the periodic table is equal to (i) magnetic quantum number of any element of the period. (ii) an atomic number of any element of the period. (iii) maximum Principal quantum number of any element of the period. (iv) maximum Azimuthal quantum number of any element of the period. Solution: Option (iii) maximum Principal quantum number of any element of the period is the answer....
Read More →Among halogens,
Question: Among halogens, the correct order of the amount of energy released in electron gain (electron gain enthalpy) is: (i) F Cl Br I (ii) F Cl Br I (iii) F Cl Br I (iv) F Cl Br I Solution: Option (iii)F Cl Br I is the answer....
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\left(a^{2}-b^{2}\right)}{c^{2}}$ Solution: Need to prove: $\frac{\sin (A-B)}{\sin (A+B)}=\frac{\left(a^{2}-b^{2}\right)}{c^{2}}$ We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius. Therefore, $a=2 R \sin A \cdots(a)$ Similarly, b = 2R sinB and c = 2R sinC From Right hand side, $=\frac{a^{2}-b^{2}}{c^{2}}$ $=\frac{4 R^{2} \sin ^{2} A-4 R^{2} \sin ^{2} B}{4 R^{2} \sin ^{2} C}$ ...
Read More →The statement that is not correct for periodic
Question: The statement that is not correct for periodic classification of elements is: (i) The properties of elements are periodic function of their atomic numbers. (ii) Non-metallic elements are less in number than metallic elements. (iii) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals. (iv) The first ionisation enthalpies of elements generally increase with increase in atomic number as we go along a period. Solution: Option (iii) Fo...
Read More →The electronic configuration
Question: The electronic configuration of gadolinium (Atomic number 64) is (i) [Xe] 4f35d56s2 (ii) [Xe] 4f75d2 6s1 (iii) [Xe] 4f75d16s2 (iv) [Xe] 4f85d66s2 Solution: Option (iii)[Xe] 4f75d16s2 is the answer....
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