The correct answer is 2
\(I_{R_1}=I_1+I_2+2\sqrt{I_1I_2}\cos\phi\)
\(I_A=I+4I+2\sqrt{I⋅4I}\cos90^∘=5I\)
\(I_B=I+4I+2\sqrt{I⋅4I}\cos60^∘=7I\)
\(I_B–I_A=2I\)
\(\therefore\) value of x is 2

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.