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Consider the following orders given below.
(i) S > Se > Te > O (Electron Affinity)
(ii) N–3 > O2– > F > Na+ > Mg2+ (Ionic Size)
(iii) Fe2+ > Fe3+ > Fe4+ (Ionization Enthalpy)
(iv) H < B < P < C < Cl (Electronegativity)
(v) Na+ < Mg2+ < Al3+ (Extent of hydration)
(vi) Li > Na > K > Rb > Cs (Reducing Nature)
(vii) LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)
If x = number of correct orders
   y = number of incorrect orders
then the value of (1 + y/x) is.
    Correct answer is '1.75'. Can you explain this answer?
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    Consider the following orders given below.(i)S > Se > Te > O ...
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    Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer?
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    Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer?.
    Solutions for Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
    Here you can find the meaning of Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer?, a detailed solution for Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? has been provided alongside types of Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider the following orders given below.(i)S > Se > Te > O (Electron Affinity)(ii)N–3> O2–> F–> Na+> Mg2+(Ionic Size)(iii)Fe2+> Fe3+> Fe4+(Ionization Enthalpy)(iv)H < B < P < C < Cl (Electronegativity)(v)Na+< Mg2+< Al3+(Extent of hydration)(vi)Li > Na > K > Rb > Cs (Reducing Nature)(vii)LiF(s) > LiCl(s) > LiBr(s) (Lattice Energy)If x = number of correct orders y = number of incorrect ordersthen the value of (1 + y/x) is.Correct answer is '1.75'. Can you explain this answer? tests, examples and also practice JEE tests.
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