The electron affinity values (in kJ mol−1) of three halogens X, Y and Z are respectively−349,−333and −325.
X, Y, and Z respectively are :
F2,Cl2,Br2
Cl2,F2 and Br2
Br2,Cl2 and F2
Cl2,Br2,F2
A
Br2,Cl2 and F2
B
F2,Cl2,Br2
C
Cl2,F2 and Br2
D
Cl2,Br2,F2
Open in App
Solution
Verified by Toppr
According to the values electron affinity the three halogens X, Y and Z are Cl2,F2 and Br2
Hence option B is correct
Was this answer helpful?
2
Similar Questions
Q1
The electron affinity values (inkJmol−1) of three halogens x, y and z are respectively -349, -333 and -325. Then x, y and z are respectively
View Solution
Q2
Consider the following orders –
(1) F2>Cl2>Br2>I2: boiling point
(2) F2>Cl2>Br2>I2: oxidizing nature
(3) F2>Cl2>Br2>I2: EN
(4) F2>Cl2>Br2>I2: BDE
(5) F2>Cl2>Br2>I2: EA
(6) F2>Cl2>Br2>I2: Reactivity
(7) HOCl>HClO2>HClO3>HClO4: Acidic nature
(8) HOCl>HClO2>HClO3>HClO4: Oxidizing nature
Then calculate (x2+y2) when x is correct order and y is incorrect order.
View Solution
Q3
The electron affinity values (in kJ mol−1) of three halogens X, Y and Z are respectively−349,−333and −325.
X, Y, and Z respectively are :
View Solution
Q4
Increasing oxidising nature of I2,Br2,Cl2,F2 is : I2<Br2<Cl2<F2. if true enter 1, else enter 0.