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Would you expect the electronegativity of titanium to be the same in the species \(\mathrm{Ti}, \mathrm{Ti}^{2+}, \mathrm{Ti}^{3+}\), and \(\mathrm{Ti}^{4+} ?\) Explain.

Short Answer

Expert verified
No, the electronegativity of titanium would not be the same in the species Ti, \(\mathrm{Ti}^{2+}\), \(\mathrm{Ti}^{3+}\), and \(\mathrm{Ti}^{4+}\). Electronegativity increases with the increasing positive charge on the titanium ion due to increasing effective nuclear charge experienced by the outermost electrons. Neutral titanium has the lowest electronegativity, while \(\mathrm{Ti}^{4+}\) has the highest electronegativity.

Step by step solution

01

Electronegativity is a measure of the ability of an atom to attract a bonding pair of electrons towards itself. It is affected by the atomic radius, the number of protons in the nucleus, and the effective nuclear charge experienced by the outermost electrons. As we move across the periodic table from left to right, electronegativity increases due to an increasing effective nuclear charge. As we move down a group, electronegativity generally decreases due to the increasing atomic radius and shielding effect. #Step 2: Discuss the effect of charge on electronegativity#

When an atom gains or loses electrons, its charge state changes. Higher positive charges generally result in greater electronegativity because there is a stronger attraction between the electrons and the nucleus. This is because the effective nuclear charge experienced by the outermost electrons increases with the loss of an electron. Conversely, a higher negative charge would generally decrease electronegativity due to the increased repulsion between the added electrons and the existing ones. #Step 3: Analyze electronegativity for different charge states of titanium#
02

Now we can analyze the electronegativity in the given titanium species: neutral titanium (Ti), \(\mathrm{Ti}^{2+}\), \(\mathrm{Ti}^{3+}\), and \(\mathrm{Ti}^{4+}\). As titanium loses electrons to form positively charged ions, the effective nuclear charge experienced by the remaining electrons increases, which makes the positively charged ions more electronegative. Therefore, we can conclude that the electronegativity would increase from neutral titanium to \(\mathrm{Ti}^{4+}\). #Step 4: Summarize the findings and answer the question#

In conclusion, the electronegativity of titanium would not be the same in the species Ti, \(\mathrm{Ti}^{2+}\), \(\mathrm{Ti}^{3+}\), and \(\mathrm{Ti}^{4+}\). Electronegativity would increase with the increasing positive charge on the titanium ion due to increasing effective nuclear charge experienced by the outermost electrons. Neutral titanium has the lowest electronegativity, while \(\mathrm{Ti}^{4+}\) has the highest electronegativity.

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: Given the following bond lengths, $$ \begin{array}{llll} \mathrm{N}-\mathrm{N} & 167 \mathrm{pm} & \mathrm{N}=\mathrm{O} & 115 \mathrm{pm} \\ \mathrm{N}=\mathrm{N} & 120 \mathrm{pm} & \mathrm{N}-\mathrm{O} & 147 \mathrm{pm} \\ \mathrm{N} \equiv \mathrm{N} & 110 \mathrm{pm} & & \end{array} $$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

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