16 public questions tagged with this topic.
Let C = x. H = +1, Cl = -1. Equation: x + 3(+1) + (-1) = 0, x + 3 - 1 = 0, x = -2.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
Ag (+1 to 0): change = -1. C (0 to +4 in CO₂): change = +4. C has the largest magnitude change.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
Let C = x. H = +1. Equation: x + 4(+1) = 0, x + 4 = 0, x = -4.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Electrode Potential and Electrochemical Series and Galvanic Cells
C in CH₄: -4. In CHCl₃: x + 1 + 3(-1) = 0, x - 2 = 0, x = +2, which is higher.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Classical and Electronic Concept of Redox and Oxidation Number
Let C = x. H = +1, Cl = -1. Equation: x + 3(+1) + (-1) = 0, x + 3 - 1 = 0, x = -2.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Classical and Electronic Concept of Redox and Oxidation Number
Let C = x. H = +1. Equation: 2x + 2(+1) = 0, 2x + 2 = 0, x = -1.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Classical and Electronic Concept of Redox and Oxidation Number
The standard state of an element is its most stable form at 1 bar and 298 K. For carbon, this is graphite.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
In HCN , carbon has 2 bonding domains (1 triple bond, 1 single bond), requiring sp hybridization, forming a linear shape.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
In CO , carbon forms a triple bond with oxygen (one sigma, two pi), using two hybrid orbitals ( sp ) for the sigma bond and lone pair.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Kossel-Lewis Approach and Octet Rule and Lewis Structures
Boiling point of group 14 tetrachlorides (XCl₄) increases with molecular weight due to stronger London forces. SnCl₄ (Sn, heaviest) has the highest boiling point among C, Si, Ge, and Sn.
Ref: NCERT Class 11 Chemistry > Chapter 3: Classification of Elements and Periodicity in Properties > Topic: Nomenclature of Elements with Atomic Number >100
C (group 14, valence 4) and Cl (group 17, valence 1) form CCl₄ by sharing electrons to achieve stable configurations.
Ref: NCERT Class 11 Chemistry > Chapter 3: Classification of Elements and Periodicity in Properties > Topic: Modern Periodic Law and Present Form - Long Form Periodic Table
C (group 14, period 2) has a diagonal relationship with Si (group 14, period 3) due to similar properties, unlike B, N, or P.
Ref: NCERT Class 11 Chemistry > Chapter 3: Classification of Elements and Periodicity in Properties > Topic: Anomalous Properties and Second Period and Chemical Reactivity
