For electron gain enthalpies of the elements denoted as\(\Delta_{ eg } H\), the incorrect option is :
Remember the general trend of electron gain enthalpy down a group. Be aware of exceptions, particularly between the second and third periods.
\(\Delta_{ eg } H ( Cl )<\Delta_{ eg } H ( F )\)
\(\Delta_{ eg } H ( I )<\Delta_{ eg } H ( At )\)
\(\Delta_{ eg } H ( Te )<\Delta_{ eg } H ( Po )\)
\(\Delta_{ eg } H ( Se )<\Delta_{ eg } H ( S )\)$\Delta_{ eg } H ( Se )<\Delta_{ eg } H ( S )$
Electron Gain Enthalpy Trend
Electron gain enthalpy generally becomes less negative (less exothermic) down a group due to increasing atomic size. However, there can be exceptions due to factors like electron-electron repulsion and shielding effects. Fluorine has a smaller atomic size and therefore greater electron density and experiences inter electronic repulsions more than chlorine and hence its magnitude is smaller than that of Cl.
Analyzing the Options
(1) Cl > F (Correct, due to small size and inter electronic repulsions of F) (2) S > Se (Incorrect, electron gain enthalpy becomes less negative down the group; therefore, S should have a more negative electron gain enthalpy than Se. This is an exception). (3) At > I (Correct) (4) Po > Te (Correct)
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Since many elements were being discovered in the 19th century and the study of these elements individually was proving difficult, classification of elements was made necessary.
Classification by Johann Dobereiner - German chemist Johann Dobereiner classified certain elements on the basis of their similar properties in the groups of continuing - three elements each. These groups were called ‘triads’. In every triad, the atomic weight of the middle element was equal to the average of the atomic weights of the first and third elements.
Newlands Law of Octaves - The elements were arranged in increasing order of their atomic weights and found that every 8th element shows similarity with the 1st element.
Mendeleev’s Periodic Table - The arrangement of all 63 elements in rows or columns in order of their atomic weight was made by Mendeleev. He left some space for corresponding elements in his periodic table which were not even discovered till then. Although he predicted the properties of those elements through his periodic classification of elements.
Modern Periodic Law - The properties of the elements of the modern periodic law are periodic functions of their atomic numbers.