How does the metallic and non-metallic
Question: How does the metallic and non-metallic character vary on moving from left to right in a period? Solution: Metallic character decreases when we move from left to right across the period and non-metallic character increases as there is an increase in ionisation enthalpy and electron gain enthalpy along the period....
Read More →Explain the following:
Question: Explain the following: (a) Electronegativity of elements increases on moving from left to right in the periodic table. (b) Ionisation enthalpy decrease in a group from top to bottom? Solution: (a) As we move from left to right in a period, the size of the atoms decreases due to the increase in the effective nuclear charges on the outermost electron. As a result of which electronegativity of elements increases on moving from left to right in the periodic table. (b) As we go down the gro...
Read More →Arrange the elements N, P, O
Question: Arrange the elements N, P, O and S in the order of- (i) increasing first ionisation enthalpy. (ii) increasing non-metallic character. Give the reason for the arrangement assigned. Solution: (i) S P O N is the increasing order of the first ionisation enthalpy. As we go down the group the ionisation enthalpy decreases and as we go along the period then it increases but in case of oxygen and nitrogen due to half-filled stability of 2 p orbitals of nitrogen it has a higher ionisation entha...
Read More →Solve this
Question: If in a $\Delta \mathrm{ABC}, \angle \mathrm{C}=90^{0}$, then prove that $\sin (\mathrm{A}-\mathrm{B})=\frac{\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}{\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)}$ Solution: Given: $\angle C=90^{\circ}$ Need to prove: $\sin (A-B)=\frac{\left(a^{2}-b^{2}\right)}{\left(a^{2}+b^{2}\right)}$ Here, $\angle C=90^{\circ} ; \sin C=1$ So, it is a Right-angled triangle. And also, $a^{2}+b^{2}=c^{2}$ Now $\frac{a^{2}+b^{2}}{a^{2}-b^{2}} \sin (A-B)=\frac{c^{2}}{a^...
Read More →What do you understand by exothermic
Question: What do you understand by exothermic reaction and endothermic reaction? Give one example of each type. Solution: Exothermic reaction- The reaction in which heat is evolved is called an exothermic reaction. for example, Cao + CO2 CaCO3 ΔH=-178kJmol-1 Endothermic reaction- The reaction in which heat is absorbed is called an endothermic reaction. for example, 2NH3 3N2+ H2 ΔH=918kJmol-1...
Read More →How would you explain the fact that first
Question: How would you explain the fact that first ionisation enthalpy of sodium is lower than that of magnesium but its second ionisation enthalpy is higher than that of magnesium? Solution: Sodium attains stable configuration when it loses on an electron from its outermost shell. That is why its first ionisation enthalpy is less than magnesium. But in case of second ionisation enthalpy magnesium has one electron in its outermost shell and to attain stability it loses that easily as compared t...
Read More →p-Block elements form acidic,
Question: p-Block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water. Solution: ACIDIC OXIDES SO2, B2O3are acidic oxides and p block elements. The reaction of SO2with water SO2+ H2O H2SO3 The reaction of B2O3 with water B2O3+3 H2O 2H3BO3 Acidic oxides are those oxides that form acids after reacting with water. BASIC OXIDES Cao, BaO, Ti2O form basic oxides The reaction of Ti2O with water. BaO + H2...
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $x=3 \cos \theta-\cos ^{3} \theta, y=3 \sin \theta-\sin ^{3} \theta$ Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to theta $\frac{\mathrm{dx}}{\mathrm{d} \theta}=-3 \sin \theta+3 \cos ^{2} \theta \sin \theta$ $\frac{\mathrm{dy}}{\mathrm{d} \theta}=3 \cos \theta-3 \sin ^{2} \theta \cos \theta$ Now dividing $\frac{d y}{d \theta}$ and $\frac{d x}{d \theta}...
Read More →The first member of each group of representative
Question: The first member of each group of representative elements (i.e., s and p-block elements) shows anomalous behaviour. Illustrate with two examples. Solution: Lithium and beryllium are examples. Li is the first group element. It has different properties and forms of covalent compounds and nitrides. Beryllium is the first element of the second group. It has various anomalies like it forms a covalent compound with coordination number four, unlike other elements that have a coordination numb...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$ Solution: Need to prove: $\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$ We know $\tan A=\frac{a b c}{R} \frac{1}{b^{2}+c^{2}-a^{2}}-\cdots-\cdots(a)$ Similarly, $\tan B=\frac{a b c}{R} \frac{1}{c^{2}+a^{2}-b^{2}}$ and $\tan C=\frac{a b c}{R} \frac{1}{a^{2}+b^{2}-c^{2}}$ Therefore,...
Read More →Nitrogen has positive electron gain enthalpy
Question: Nitrogen has positive electron gain enthalpy whereas oxygen has negative. However, oxygen has lower ionisation enthalpy than nitrogen. Explain. Solution: The ionisation enthalpy of oxygen is lower than that of Nitrogen as because when we remove one electron from oxygen then it easily donates it to attain half-filled stability but in case of nitrogen, it is difficult to remove one electron as it already has half-filled stability and it would become unstable after that....
Read More →Illustrate by taking examples of transition
Question: Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration. Solution: Ti has an atomic number of 22 and electronic configuration [Ar]3d24s2 and can show three oxidation states of +2,+3 and +4 in various compounds like TiO29(+4), Ti2O3(+3) and TiO(+2). The non-transition elements like the p block elements show variable oxidation states like in case of phosphorous. It has -3,+3 and +5....
Read More →Choose the correct order of atomic radii
Question: Choose the correct order of atomic radii of fluorine and neon (in pm) out of the options given below and justify your answer. (i) 72, 160 (ii) 160, 160 (iii) 72, 72 (iv) 160, 72 Solution: (i) 72, 160 Because neon has van der Waals radii and fluorine has covalent radii. Covalent radius is always less than van der Waals radius, so the radius of Fluorine is 72pm and Neon is 160pm....
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}=\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)$ Solution: Need to prove: $\frac{\cos 2 \mathrm{~A}}{\mathrm{a}^{2}}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^{2}}=\left(\frac{1}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}\right)$ Left hand side, $=\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}$ $=\frac{1-2 \sin ^{2} A}{a^{2}}-\frac{1-2 \sin ^{2} B}{b^{2}}$ $=\frac{1}{a^{2}}-\frac{1}{b^{2}}+2\left(\frac{\sin ^{2} B}{b^{2}}-\f...
Read More →Write four characteristic properties
Question: Write four characteristic properties of p-block elements. Solution: 1. They show variable oxidation states. The reducing character increases down the group and oxidising character increases along the period. 2. They have a high ionisation enthalpy than the s-block elements. 3. They usually form covalent compounds. 4. Both metals and non-metals are found in this group, but non-metals are slightly more in number....
Read More →Among the elements B, Al, C and Si,
Question: Among the elements B, Al, C and Si, (i) which element has the highest first ionisation enthalpy? (ii) which element has the most metallic character? Justify your answer in each case. Solution: (i) Carbon has the highest ionisation enthalpy. It increases from left to right across the period and decreases as we go down the group. (ii) Aluminium has the most metallic character. As we move down, the metallic character increases and decreases across the period from left to right....
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $\frac{\left(\cos ^{2} \mathrm{~B}-\cos ^{2} \mathrm{C}\right)}{\mathrm{b}+\mathrm{c}}+\frac{\left(\cos ^{2} \mathrm{C}-\cos ^{2} \mathrm{~A}\right)}{\mathrm{c}+\mathrm{a}}+\frac{\left(\cos ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~B}\right)}{\mathrm{a}+\mathrm{b}}=0$ Solution: Need to prove: $\frac{\left(\cos ^{2} \mathrm{~B}-\cos ^{2} \mathrm{C}\right)}{\mathrm{b}+\mathrm{c}}+\frac{\left(\cos ^{2} \mathrm{C}-\cos ^{2} \mathrm{~A}\right)}{\mathrm{c}+\mathrm{a}}+\frac...
Read More →Identify the group and valency of the element
Question: Identify the group and valency of the element having atomic number 119. Also, predict the outermost electronic configuration and write the general formula of its oxide. Solution: There are 118 elements in the 7 periods of the modern periodic table. Therefore the element with atomic number 119 will lie in the 8th period of the first group and will have the outermost electronic configuration of 8s1. It belongs to group 1 and has a valency of one. The formula of its oxide will be M2O....
Read More →Find the equation
Question: Find the equation of the tangent and the normal to the following curves at the indicated points: $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $\theta$ Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to theta $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)$ $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(\sin \theta)$ Now dividing $\frac{d y}{d \theta}$ and $\frac{d x}{d \theta}$ to obtain the slope of tangent $\frac{d y}{d x}=\f...
Read More →All transition elements are d-block elements,
Question: All transition elements are d-block elements, but all d-block elements do not transition elements. Explain. Solution: The elements having their outermost shell filled with d electrons are known as d block elements. All d block are not transition elements because it is important to have incompletely filled d orbital of the element like calcium and zinc etc....
Read More →Explain why the electron gain enthalpy
Question: Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine. Solution: Fluorine has a smaller size as compared to chlorine as a result of which the attraction outside the shell to gain electron is less. Moreover, there are inter electronic repulsions as well in the 2p orbitals which results in the less negative electron gain enthalpy....
Read More →An element belongs to the 3rd period
Question: An element belongs to the 3rd period and group-13 of the periodic table. Which of the following properties will be shown by the element? (i) Good conductor of electricity (ii) Liquid, metallic (iii) Solid, metallic (iv) Solid, non-metallic Solution: Option (i)Good conductor of electricity and (iii) Solid, metallicare 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: $x=a \sec t, y=b \tan t a t t$ Solution: finding slope of the tangent by differentiating $x$ and $y$ with respect to $t$ $\frac{\mathrm{dx}}{\mathrm{dt}}=\operatorname{asecttan} \mathrm{t}$ $\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{bsec}^{2} \mathrm{t}$ Now dividing $\frac{d y}{d t}$ and $\frac{d x}{d t}$ to obtain the slope of tangent $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b} \opera...
Read More →In any ΔABC, prove that
Question: In any ΔABC, prove that $a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=0$ Solution: Need to prove: $a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=$ 0 From left hand side, $=a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)$ $=a^{2}\left(\left(1-\sin ^{2}...
Read More →Ionic radii vary in
Question: Ionic radii vary in (i) inverse proportion to the effective nuclear charge. (ii) inverse proportion to the square of effective nuclear charge. (iii) direct proportion to the screening effect. (iv) direct proportion to the square of screening effect. Solution: Option (i) and (iii) are the answers....
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